Functions of Several Variables, Limits, Partial Derivatives & Differentiability
A scalar-valued function of several variables maps geometric coordinate points to real numbers — modeling physical quantities such as gravitational potential, atmospheric temperature fields, or economic utility landscapes. While single-variable calculus considers motion along a one-dimensional line, multivariable calculus reveals how a multidimensional surface tilts, curves, and deforms in every conceivable direction. The foundational building blocks of this domain are coordinate traces, multivariable limits, partial derivatives, and the rigorous standard of Fréchet differentiability.
A function with domain assigns a unique real number to each ordered pair in . Its graph is the surface .
The Two-Path Test for Non-Existence of Limits: We write if for every , there exists such that . Unlike single-variable limits where can only approach from two directions (left and right), in there are infinitely many trajectories approaching .
Theorem (Two-Path Test): If approaches two different values along two distinct paths approaching , or if the limit along any single path depends on a direction parameter (such as slope ), then:
Fixing restricts the surface to the vertical plane , generating the 1D trace curve . The slope of the tangent line to at is the partial derivative with respect to :
Similarly, fixing yields the trace curve in the plane , whose slope at is the partial derivative with respect to :
Operational Rule: To compute , treat as a constant and differentiate with respect to using standard single-variable calculus rules. To compute , treat as a constant.
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f(x, y) = 5 − 0.3x² − 0.2y² + 0.1xy. Fixing y = c gives the trace g(x) = f(x, c) from Definition 1.1; its slope at x = 1 is the partial fₓ(1, c) from Definition 1.2, and it shifts as c moves.
Evaluate or prove that it does not exist.
Step 1: Test straight lines : Along the coordinate axes and , the value is also identically 0. A student might prematurely conclude the limit is 0!
Step 2: Test parabolic paths :
Notice that in the denominator, and have matching degree when . Choose the path : Along (), the limit is . Along (), the limit is .
Conclusion: Because the limit depends on the choice of path , by the Two-Path Test, the limit does not exist (DNE).
Tangent Plane to a Surface: Let have continuous partial derivatives at . The tangent vectors to the coordinate traces at (where ) are and . Their cross product produces the normal vector:
The equation of the tangent plane to at is:
The Linear Approximation & Total Differential: The linear approximation (tangent plane function) is . For infinitesimal increments and , the total differential is:
Let with open domain . If the mixed second-order partial derivatives and both exist on and are continuous at , then:
Fix . For small nonzero increments such that the rectangle , define the second-order difference quotient:
Perspective 1 ( then ): Define the auxiliary single-variable function . Then: Since exists, is differentiable with derivative . By the Mean Value Theorem on , there exists such that: Now define . Applying the Mean Value Theorem to on , there exists such that: Substituting back:
Perspective 2 ( then ): Symmetrically, define . Then . Applying the Mean Value Theorem first to on (yielding ) and then to on (yielding ):
Equating perspectives and taking limits: Since both expressions equal : As , and . By continuity of and at :
A closed cylindrical storage tank has nominal measured radius and height . The measurement errors are bounded by and .
- Find the tangent plane linear approximation for the volume at .
- Use the total differential to estimate the maximum propagated error in the calculated volume, and compute the relative percentage error.
Step 1: Compute partial derivatives and nominal volume: Tangent plane: .
Step 2: Total differential & maximum error bound: Applying the triangle inequality for worst-case error:
Step 3: Relative and percentage error: (Notice that radius error contributes twice as heavily as height error due to the quadratic exponent !)
In single-variable calculus, existence of is equivalent to differentiability. In multivariable calculus, the existence of all partial derivatives does NOT imply that a function is differentiable (or even continuous)!
Definition (Differentiability): Let and let be an interior point of . We say is differentiable at if both partial derivatives and exist, and: Equivalently, the error satisfies as .
Theorem: If the first-order partial derivatives and exist in an open neighborhood of and are continuous at (i.e., ), then is differentiable at .
Hierarchy of Smoothness:
Let . Decompose this increment into two single-variable steps: Applying the single-variable Mean Value Theorem to each bracketed difference:
- To on : there exists such that:
- To on : there exists such that:
Now rewrite: where and .
As , and . Because and are continuous at , and .
Dividing by : Thus the Fréchet differentiability limit is zero, proving is differentiable at .
Consider the classic pathological function:
- Compute the partial derivatives and directly from Definition 1.2.
- Prove that is discontinuous at and therefore not differentiable at the origin.
Step 1: Compute partials from the limit definition: By symmetry: Both partial derivatives exist cleanly and equal 0 at the origin!
Step 2: Test continuity at the origin:
Approach along the line : Along (), . Along (), . Along the axes, .
Since does not exist, is discontinuous at . Because differentiability implies continuity, is not differentiable at the origin despite having well-defined partial derivatives there! (Why does this not violate Theorem 1.2? Because and are discontinuous at .)
- The Straight-Line Limit Trap: Showing that along every straight line does NOT prove the limit exists! You must test degree-matching polynomial curves (such as parabolas ) or prove the limit rigorously using the Squeeze Theorem or polar coordinates ( with ).
- Existence of Partials ≠ Continuity: As shown in Worked Example 1.C, having both and defined at a point does NOT guarantee that the surface is continuous or smooth there. Differentiability requires the tangent plane to approximate the surface to first-order in all radial directions simultaneously.
- Treating Inactive Variables as Zero: When computing , novice students mistakenly substitute . The factor is a constant multiplier; the derivative is , never 0.
- Tangent Plane Normal Vector Sign: For , the upward normal vector is and the downward normal is . Writing has the wrong signs for both coordinate slopes.
Core Multivariable Derivative Rules
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Step 1: Differentiate with respect to (holding constant):
Step 2: Differentiate with respect to (holding constant):
Step 3: Evaluate :
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University Multivariable Rigor: The tangent plane is the best linear approximation to the surface near $(x_0, y_0)$. The upward surface normal is orthogonal to every tangent vector on the surface!العمق الجامعي في حساب التفاضل متعدد المتغيرات: المستوى المماس هو أفضل تقريب خطي للسطح بالقرب من $(x_0, y_0)$. المتجه العمودي العلوي على السطح متعامد مع كل متجه مماس على السطح!
Step 1: Compute partial derivatives:
Step 2: Evaluate at :
Step 3: Apply the tangent plane formula:
The Multivariable Chain Rule, Directional Derivatives & Gradient Geometry
While Chapter 1 isolated rates of change along standard coordinate axes, physical processes rarely restrict their motion to orthogonal grid lines. A heated particle moves along a curved parametric path; fluid flows across an arbitrary streamline; optimization algorithms seek the direction of steepest descent. The multivariable chain rule tracks how composite dependencies propagate through intermediate variables, while the gradient vector encodes the directional derivative in every possible direction, establishing the fundamental geometric bridge between linear algebra, level sets, and differential geometry.
Let be a differentiable function of and .
- Case 1 (Parametric Curve Domain): If and are differentiable functions of a single parameter , then is a differentiable function of , and its derivative is given by the tree-branch summation:
- Case 2 (Two Intermediate, Two Independent Variables): If and are differentiable functions of and , then:
- Jacobian Matrix Formulation: In general matrix terms, if and , the derivative of the composite mapping is the matrix product of their Jacobian matrices:
Let be a differentiable function where and . Express and in terms of and , and show that:
Step 1: Compute partial derivatives of coordinate functions:
Step 2: Apply the Chain Rule:
Step 3: Square and add: Adding the two lines eliminates the cross-term:
The Gradient Vector: The gradient of a scalar field at is the vector field: In , .
The Directional Derivative: Let be a unit vector (). The directional derivative of in the direction of at is: If is differentiable, the directional derivative is calculated as the inner product of the gradient with the unit direction vector:
Let be a differentiable function at with . Then:
- The maximum value of the directional derivative is , and it occurs when points in the direction of the gradient:
- The minimum value of is , and it occurs when points in the direction opposite to the gradient:
- The directional derivative is zero () in any direction orthogonal to the gradient (), which is the direction tangent to the level curve of .
By Definition 2.2, . Using the geometric definition of the inner product: where is the angle between and the unit vector (since ).
Because for all :
- achieves its maximum when . Thus is parallel and codirectional with , and .
- achieves its minimum when . Thus points opposite to , and .
- , meaning .
For at , find:
- The directional derivative toward the point .
- The maximum rate of increase at and the unit direction vector in which it occurs.
- The unit direction vectors along which the rate of change of is zero.
Step 1: Compute gradient at :
Step 2: Direction toward : Displacement vector: .
Normalize : .
Compute directional derivative:
Step 3: Maximum rate of increase: Direction: .
Step 4: Zero rate of change: Requires . Swapping coordinates with sign flip gives:
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f(x, y) = x² + xy, the same function as Worked example 2.B. The amber arrow is the fixed gradient ∇f(1,2) = ⟨4, 1⟩, which points at θ ≈ 14.04°; the indigo arrow is the unit direction u you're rotating. Drag θ near there (the whole-degree slider can only approach it, not land on it exactly) and watch D_u f = ∇f·u climb toward its true maximum √17 ≈ 4.123 — reached exactly only when u lines up with ∇f, Theorem 2.1's claim.
Let be a level surface of a continuously differentiable function . Let be a point on where .
The tangent plane to at is the plane passing through with normal vector : The normal line to at has symmetric equations:
Theorem: Let be differentiable. At any point on the level surface , the gradient vector is strictly orthogonal to every smooth curve on passing through .
Let be any smooth curve on the level surface parameterized by for , such that .
Because the curve lies entirely on the level surface , the function value remains constant along : Differentiating both sides with respect to using the multivariable chain rule (Definition 2.1): Expanding the left-hand derivative: Expressed in vector notation: Evaluating at : Since represents the tangent vector to the curve at , and was chosen arbitrarily among all smooth curves on through , the gradient vector is perpendicular to every tangent vector on . Hence, is the normal vector to the level surface.
Find the equations of the tangent plane and the normal line to the hyperboloid of two sheets at the point .
Step 1: Define implicit level function:
Step 2: Compute gradient vector: At : We can scale the normal vector by to simplify: .
Step 3: Construct tangent plane equation:
Step 4: Construct normal line equations: In parametric form: , , . In symmetric form:
- Failure to Normalize the Direction Vector: The directional derivative formula assumes is a unit vector (). If given a direction vector , taking erroneously multiplies the true rate of change by . Always compute first!
- The Missing Negative Sign in Implicit Differentiation: By the Chain Rule on : . Forgetting the minus sign is one of the most persistent exam errors in multivariable calculus.
- Level Curve Normal vs. Graph Normal: For a 2D surface , rewrite it in implicit level surface form . Its 3D normal vector is . Confusing (a 2D vector in the plane) with the 3D normal vector causes severe dimensional mismatch errors.
- Chain Rule Tree Branch Omission: When an intermediate variable appears in multiple positions (e.g. where ), appears in both and . The derivative must sum both branches: .
Essential Gradient & Chain Rule Formulations
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Step 1: Compute partial derivatives of :
Step 2: Compute derivatives of coordinate functions:
Step 3: Evaluate at :
Step 4: Apply the chain rule:
Step 1: Compute the gradient :
Step 2: Evaluate at :
Step 3: Normalize direction vector :
Step 4: Compute dot product:
Multivariable Optimization: Critical Points, Hessians & Lagrange Multipliers
Finding optimal configurations — whether minimizing the energy of a physical lattice, maximizing the profit of a multi-product firm, or training high-dimensional neural network loss landscapes — is the ultimate application of differential calculus. In several variables, extrema occur where the gradient vector vanishes, creating horizontal tangent planes. Distinguishing peaks, valleys, and saddle points requires analyzing the curvature via the Hessian matrix, while optimizing subject to real-world boundary constraints is governed by the method of Lagrange multipliers.
Let be defined on an open set .
- has a local maximum at if for all in an open disk centered at .
- has a local minimum at if for all in an open disk centered at .
- A point in is a critical point (or stationary point) of if: or if one or both partial derivatives fail to exist at .
- Fermat's Theorem for Multivariable Functions: If has a local extremum at and is differentiable there, then . (Proof: If , moving in the direction of increases and moving opposite decreases , contradicting extremality.)
- A critical point where that is neither a local maximum nor a local minimum is called a saddle point.
Find all critical points of the surface .
Step 1: Compute first-order partial derivatives and set to zero:
Step 2: Form candidate coordinate pairs:
Pairing each solution for with yields two critical points:
Function values: , and .
Let be a twice continuously differentiable function (). The Hessian matrix of at is the symmetric matrix of second-order partial derivatives: By Clairaut's Theorem (Theorem 1.1), , so is real symmetric. The Hessian discriminant (or determinant) is:
Let be a critical point of () with continuous second partial derivatives on an open disk centered at . Let .
- Local Minimum: If and (equivalently is positive definite, eigenvalues ), then is a local minimum.
- Local Maximum: If and (equivalently is negative definite, eigenvalues ), then is a local maximum.
- Saddle Point: If (equivalently is indefinite, eigenvalues have opposite signs ), then is a saddle point.
- Inconclusive: If , the test yields no information; higher-order terms determine the behavior.
Let and . By Taylor's Theorem in two variables centered at the critical point where : where is the second-order quadratic form and as . The local behavior is determined entirely by the sign of .
Case 1 (): Complete the square in : Combining the second term over a common denominator: Thus:
- If and : Both coefficients and are strictly positive. Since the sum of two squares is strictly positive for all , , confirming is a local minimum.
- If and : Both coefficients are strictly negative. Hence for all , confirming a local maximum.
- If : The coefficients have opposite signs. Setting gives , which has the sign of . Along the line , , which has the opposite sign! Because takes both positive and negative values in every neighborhood of the origin, is a saddle point.
Case 2 (): If , then . If (meaning ), then , which changes sign across the lines and , again yielding a saddle point.
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f(x, y) = x³ − 3x + y², the same function as Worked example 3.A. The indigo dot at (1, 0) and the amber dot at (−1, 0) are its two named critical points. Drag (a, b) with the ring marker: f_x and f_y are computed at every point, but Theorem 3.1's D/f_xx classification only applies where ∇f = 0 — move off (±1, 0) and the readout correctly reports 'Not a critical point' instead of guessing a type.
When optimizing subject to a constraint equation , candidate points are restricted to the level curve .
Theorem (Lagrange Multipliers): Suppose and have continuous first partial derivatives, and is an extremum of restricted to . If , then there exists a scalar (the Lagrange multiplier) such that: In coordinate equations, this system plus the constraint yields 3 equations in 3 unknowns :
Two Constraints in : To optimize subject to two simultaneous constraints and (whose intersection forms a space curve), the gradient must lie in the plane spanned by and :
Let be any regular smooth parameterization of the constraint curve near , with and tangent vector .
Along the constraint curve, is identically constant. Differentiating with respect to via the Chain Rule: Thus, is perpendicular to the tangent vector .
Now consider the objective function along the curve: . Since is a local extremum of restricted to the curve, has an ordinary single-variable extremum at , so Fermat's Theorem requires . Applying the Chain Rule:
In the two-dimensional plane , the orthogonal complement of the 1D tangent line spanned by is a one-dimensional subspace. Because both and are perpendicular to , they must be linearly dependent (collinear). Since , there exists a scalar such that .
Find the absolute maximum and minimum values of on the closed triangular region bounded by the lines , , and .
Step 1: Critical points in the interior of : The interior point lies inside since , and .
Value: .
Step 2: Boundary Segment 1 ( for ):
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. Candidate points: , , .
Values: ; ; .
Step 3: Boundary Segment 2 ( for ):
.
. Candidate points: , .
Values: ; .
Step 4: Boundary Segment 3 ( for ):
.
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Value: .
Step 5: Global comparison: Comparing all candidate values :
- Discriminant Sign Priority: Always check the sign of FIRST. If , the point is immediately a saddle point, regardless of whether is positive, negative, or zero. Looking at before verifying is a classic exam blunder.
- Inconclusive Test : If , the test fails completely. For example, has and has a local minimum, while also has but is a saddle point. You must analyze the direct behavior of the function along test lines or parabolas.
- Omitting Boundary Search on Compact Sets: When finding absolute extrema on a closed, bounded domain (e.g., a disk or triangle), interior critical points alone are not enough! The absolute maximum or minimum often lies on the boundary curve .
- Singular Points on Constraints: In Lagrange multipliers, solutions can also occur where (points where the constraint curve is not smooth or has a cusp). These singular points must be checked independently alongside .
Optimization & Classification Summary
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Step 1: Set first partial derivatives equal to zero:
Step 2: Solve the system: From , either or .
- If : or . Critical points: and .
- If : . Critical points: and .
Step 3: Compute second-order partials:
Step 4: Classify each point:
- At : and .
- At : , and .
- At : , .
- At : .
Step 1: Set up the Lagrange equations: Constraint . This gives:
Step 2: Substitute into constraint:
Step 3: Find corresponding points:
- For : , .
- For : , .
Step 4: Evaluate :
Double Integrals, Fubini's Theorem, Polar Coordinates & The Change-of-Variables Jacobian
While a single definite integral accumulates a function over a one-dimensional interval, a double integral computes the continuous accumulation of a scalar field over a two-dimensional planar domain — measuring volumes under surfaces, mass distributions with variable areal density, and probability densities. By virtue of Fubini's Theorem, 2D continuous integrals decompose into iterated single-variable integrations, while curvilinear transformations are governed by the Jacobian determinant.
Let be a continuous function on a closed rectangle . Partition into subintervals of width and into subintervals of width , forming rectangular sub-boxes each of area . In each sub-rectangle, choose a sample point .
The double integral of over is the limit of the double Riemann sums as grid dimensions shrink to zero: When , this limit represents the exact volume of the 3D solid bounded between the surface and the -plane over .
If is continuous on the rectangle , then the double integral equals the iterated integrals in either order:
Separable Integrand Case: If , the double integral splits into the direct product of two independent single-variable integrals:
Let on . For any partition of into sub-rectangles with sample points , the double Riemann sum factors algebraically by the distributive law of real numbers: As , because and are continuous on compact intervals, both bracketed factors are standard single-variable Riemann sums that converge independently to and . Because the limit of a product of convergent sequences is the product of their limits: Evaluating the iterated integral: , identical to the product. Reversing the order yields the same result by symmetry.
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Each cell is shaded by f(x,y) = x² + y at its midpoint over [0,2]×[0,1] — the same integral as Worked example 4.A. Refine the grid and the midpoint sum converges to 11/3 ≈ 3.667.
Double integrals extend to non-rectangular, bounded regions :
- Type I (Vertically Simple): . The integral is evaluated by integrating first:
- Type II (Horizontally Simple): . The integral is evaluated by integrating first:
- Reversing the Order of Integration: If an inner anti-derivative is non-elementary (e.g. , , ), sketching the geometric boundary curves allows converting the region between Type I and Type II representations, making the integral elementary.
Evaluate the integral by reversing the order of integration:
Step 1: Identify and sketch region :
The current iterated bounds are: for .
This describes a triangle in the -plane bounded by the lines , , and .
Step 2: Express as a Type I region:
For each fixed , ranges from the bottom boundary up to the line :
Step 3: Evaluate the reversed iterated integral: Substitute . When , ; when , :
Let be a bijective coordinate transformation given by , that maps a region in the -plane onto in the -plane.
The Jacobian determinant of the transformation is:
Geometric Derivation: A small coordinate rectangle in the -plane with sides and maps to a curvilinear parallelogram in the -plane spanned by tangent vectors: The area of this mapped parallelogram is the magnitude of their cross product:
Theorem: Under the smooth, invertible transformation :
Polar Coordinates Transformation: Setting and : Since , the absolute value is , establishing the canonical area element:
Prove rigorously that the Gaussian integral satisfies:
Proof: The single-variable anti-derivative cannot be expressed in terms of elementary functions. However, write: Since the variables and are independent dummy integration variables, multiplying the two integrals yields a double integral over the entire plane by Fubini's Theorem (Theorem 4.1):
Convert the double integral to polar coordinates where , , , and : Compute the -integral using substitution : Multiplying by the angular integral: Taking the positive square root (since ):
- The Omitted Factor of in Polar Coordinates: Writing is the single most common error in integral calculus. Due to the area expansion of polar sectors, . Neglecting yields invalid dimensions and incorrect numerical integrals.
- Variables in Outer Limits: In any iterated integral , the outer limits and must strictly be constants. An outer integral like is mathematically meaningless and indicates a fundamental failure in reversing bounds.
- Jacobian Absolute Value: In the change of variables formula, you must take the absolute value of the Jacobian determinant: . If a coordinate transformation reverses orientation (producing ), omitting the absolute value yields an incorrect negative area element.
- Separability Fallacy: You may only split if BOTH the integrand is separable () AND the region is a Cartesian rectangle with constant limits! If the limits depend on or , factoring across integrals is illegal.
Double Integration Principles
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Step 1: Sketch and identify the region : The given bounds are for . This is the triangle bounded by , , and .
Step 2: Reverse order to integrate with respect to first: In the reversed order, for :
Step 3: Evaluate inner integral:
Step 4: Evaluate outer integral using substitution , :
Step 1: Convert region to polar coordinates: The upper half-disk of radius 3 corresponds to and .
Step 2: Convert integrand and area element:
Step 3: Set up and evaluate iterated integral:
Triple Integrals & Cylindrical/Spherical Coordinates
A triple integral extends multivariable integration to three-dimensional solid regions . When the integrand is the unit function , the integral computes the geometric volume ; when represents density, it yields the total mass , centroid coordinates , and moments of inertia . As in double integrals, Fubini's Theorem reduces the volume integral to iterated single-variable integrals. Mastery of this domain requires navigating coordinate transformations—principally cylindrical and spherical coordinate systems—governed by the 3D Jacobian determinant.
This chapter develops through three instructional tiers:
- Level 1 (Foundations): Riemann sums over spatial rectangular boxes, Fubini's Theorem in , classification of elementary solids (Types I, II, III), and standard Cartesian bounds setup.
- Level 2 (Medium / Analytical Application): Cylindrical and spherical coordinate transformations, geometric derivations of differential volume elements and , center of mass, and rotational inertia.
- Level 3 (Advanced / University Rigor): The General 3D Change of Variables Theorem via the Jacobian determinant, formal derivation of , Newton's Shell Theorem for gravitational potential, and inertia tensor derivations.
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Let . Partition each interval into subintervals of widths , , and . The sub-box volume is . For any continuous scalar field , the triple integral is defined as the limit of the triple Riemann sum: provided this limit exists independently of the choice of sample points .
Statement: If is continuous on the rectangular box , then the triple integral is equal to the iterated integral evaluated in any of the possible integration orders: For a general bounded solid of Type I (projectable onto a region in the -plane between surfaces and ):
Problem: Evaluate , where is the solid tetrahedron bounded by the coordinate planes and the plane .
Analytical Solution:
- Determine the bounding surfaces: In the vertical -direction, is bounded below by and above by .
- Project onto the -plane: Setting yields the planar triangle .
- Set up the iterated integral:
- Evaluate the innermost integral:
- Integrate with respect to : Using substitution , :
- Integrate with respect to :
Thus, .
Cylindrical coordinates represent a point by , where are polar coordinates of the orthogonal projection of onto the -plane, and is the directed vertical height: The differential volume element represents a small cylindrical wedge of radial thickness , angular width (arc length ), and height : Cylindrical coordinates are natural whenever the boundary surfaces exhibit rotational symmetry about the -axis (cylinders , vertical cones , paraboloids ).
Spherical coordinates locate by , where:
- is the radial distance from the origin.
- is the colatitude (zenith angle) measured downward from the positive -axis.
- is the azimuthal angle in the -plane from the positive -axis.
Theorem: Under the transformation , the volume element scales by the absolute value of the Jacobian determinant:
Rigorous Proof: The Jacobian matrix of partial derivatives is: Expanding along the third row (which contains a zero entry): Evaluate the first minor: Evaluate the second minor: Substituting both minors into the cofactor expansion: Since , , and , .
Problem: Find the volume of the solid bounded above by the sphere and below by the cone .
Analytical Solution:
- Express the bounding surfaces in spherical coordinates:
- The sphere becomes .
- The cone .
- Identify the parameter limits: The solid is described by:
- Set up and evaluate the triple integral: Since the bounds are constants and the integrand separates:
- Calculate the individual factors:
- Multiply results:
Problem: A solid ball of radius has uniform mass density . Derive from first principles the moment of inertia about the -axis:
Analytical Derivation:
- Convert the integrand to spherical coordinates: The distance squared from the -axis is:
- Set up the integral:
- Compute the and integrals:
- Evaluate the integral using : Let , :
- Assemble the product:
- Substitute :
This yields the classical university physics result purely via multivariable integration.
1. Missing the Differential Metric Factors: Never write or . The Jacobian determinants dictate that and . Omitting or invalidates dimensional analysis and gives incorrect answers.
2. Colatitude vs. Azimuth Domain Restrictions: The angle measures deviation from the zenith (North pole) and satisfies . Integrating over double-covers the sphere and produces zero due to .
3. Variable Dependency in Limits: In an iterated integral , the limits for can depend on and ; the limits for can depend only on ; the outer limits must be constants. Outer limits containing inner variables indicate an inverted integration setup.
4. Shifted Spheres: For a sphere tangent to the -plane , the spherical equation is . The domain is restricted to (the upper hemisphere), since .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
In spherical coordinates, and . The ball is defined by , , . Evaluating each integral:
Using cylindrical coordinates, the cylinder is bounded by , , . The differential volume is . The total mass is:
Vector Fields & Line Integrals
A vector field assigns a vector—representing force, velocity, fluid flux, or a field gradient—to every point in a region of space. Integrating a vector field along an oriented space curve answers fundamental physical and geometric questions: the mechanical work performed by a force, circulation of a fluid along a loop, or magnetic flux. Central to multivariable analysis is distinguishing general vector fields from conservative (gradient) fields, where path independence allows line integrals to be evaluated purely from boundary values.
This chapter is organized into three instructional tiers:
- Level 1 (Foundations): Scalar line integrals versus vector line integrals , parametric evaluation along smooth curves, orientation reversal, and physical work.
- Level 2 (Medium / Analytical Application): Conservative vector fields , the Fundamental Theorem of Line Integrals and its proof, differential forms notation , and algorithmic reconstruction of 2D and 3D scalar potentials.
- Level 3 (Advanced / University Rigor): The three equivalent characterizations of conservatism, Clairaut's cross-partial test, topological requirements (simply connected domains), and the pathological vortex field on the punctured plane .
A vector field on () is a mapping . In components, in 2D, or in 3D.
For a piecewise smooth curve parametrized by for :
- Scalar Line Integral (with respect to arc length):
- Vector Line Integral (Work / Tangential circulation):
Reversing curve orientation negates the vector line integral: , while scalar line integrals are orientation-invariant.
Problem: Evaluate the work done by the force field in moving a particle from to along two different trajectories:
- : The straight line segment , .
- : The parabolic arc , .
Analytical Solution:
Path : Velocity . Along , and , so .
Path : Velocity . Along , and , so .
The identical values suggest that may be path-independent (conservative).
A vector field defined on an open domain is called conservative if there exists a continuously differentiable scalar field such that: The function is called a scalar potential for . (In physics, the potential energy is often defined with a negative sign: .)
Statement: Let be a piecewise smooth curve parametrized by for , with initial point and terminal point . If is a scalar function on an open domain containing , then:
Rigorous Proof: Parametrize by on . By Definition 6.1: Define the composite single-variable scalar function . By the Multivariable Chain Rule (Theorem 2.1): Applying the ordinary Fundamental Theorem of Calculus to on :
Statement: Let be a vector field on an open domain . If is conservative, then: In , if is conservative, its curl must vanish identically:
Proof: If , then and . Differentiating yields: By Clairaut's Theorem (Theorem 1.1), mixed second partial derivatives of functions commute: . Thus .
Problem: Show that is conservative on , construct a scalar potential , and evaluate along any path from to .
Analytical Solution:
- Test cross-partials:
- , .
- , .
- , .
- Integrate with respect to :
- Differentiate with respect to and equate to : Thus .
- Differentiate with respect to and equate to : Setting , the potential is .
- Evaluate line integral using Theorem 6.1:
Theorem (Three Equivalent Statements): Let be a continuous vector field on an open, connected domain . The following three conditions are logically equivalent:
- is conservative; that is, for some potential on .
- is path-independent in (depends only on the endpoints of ).
- for every piecewise smooth closed curve .
Proof of (2) (3): Let and be two paths in sharing initial point and terminal point . The composite curve (formed by traversing forward and backward) is a closed loop. Then: Therefore, .
Topological Requirement (Simple Connectedness): The component condition guarantees conservatism if and only if the domain is simply connected (every closed curve in can be continuously contracted to a point within , meaning has no "holes").
Problem: Consider the vortex vector field defined on the punctured plane : Prove that everywhere on , yet is NOT conservative on .
Analytical Demonstration:
- Compute partial derivatives: With and : Thus at every single point .
- Integrate around the counterclockwise unit circle , : Along , , , and .
- Conclusion: If were conservative on , the closed loop integral would equal by Theorem 6.3. Since , is strictly non-conservative on . The reason: is not simply connected. The candidate potential increases by each cycle and cannot be made single-valued and continuous across any loop enclosing the origin.
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The default a = 0, b = 1, c = −1, d = 0 gives F = (y, −x), a pure rotation field — every arrow is perpendicular to the position vector, circulating around the origin with nothing pulling straight outward or inward. Set b = c and the field becomes conservative (Theorem 6.2, sufficient here since F is defined on all of ℝ², which is simply connected); the readout then also displays a genuine potential function f(x,y) = (a/2)x² + bxy + (d/2)y² satisfying ∇f = F.
1. Confusing Scalar with Vector : requires multiplying by speed and is always scalar; uses the velocity vector directly via dot product without the Euclidean norm factor.
2. Forgetting the Simply Connected Requirement: The cross-partial condition only guarantees that is conservative if the domain has no holes. If the domain contains punctures (e.g. dividing by at the origin), paths that loop around the puncture can have nonzero circulation.
3. Potential Function Constant of Integration in Several Variables: When integrating with respect to , the constant of integration is a function of the remaining variables , not a numerical constant. Failing to include prevents proper matching with and .
4. Orientation Sign Flip: Traversing a curve in the opposite direction negates the value of a vector line integral: .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
First check if is conservative: Notice while . Thus is not conservative and the Fundamental Theorem cannot be used directly.
Evaluate directly: . . Regroup terms: Alternatively, evaluate . The remaining -terms integrate to: Substituting , : the integral yields . Combined work: .
Since is explicitly given as the gradient of the potential function , by the Fundamental Theorem of Line Integrals: Initial point: .
Terminal point: .
Evaluating at endpoints: Thus, .
Green's Theorem
Green's Theorem establishes a profound duality between the boundary behavior of a vector field along a closed planar loop and the interior differential behavior across the enclosed region . It represents the two-dimensional specialization of the generalized Stokes' Theorem on differential forms, bridging single-variable calculus and multivariable field theory. Green's Theorem exists in two dual formulations: the tangential / circulation form (connecting circulation to microscopic curl) and the normal / flux form (connecting outward flux to microscopic divergence).
This chapter progresses through three instructional tiers:
- Level 1 (Foundations): Positively oriented simple closed curves, the classical tangential/circulation form of Green's Theorem, and verification on elementary polygons.
- Level 2 (Medium / Analytical Application): The 2D Divergence (normal flux) Theorem, coordinate-free vector interpretations, and boundary area formulas applied to classical curves (hypocycloids, astroids).
- Level 3 (Advanced / University Rigor): Rigorous analytical proof for Type I and Type II regions, extension to multiply connected domains (annular regions with holes), and the puncture-annulus deformation argument explaining the circulation of the vortex field.
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Let be a bounded open region whose boundary consists of finitely many piecewise smooth, simple closed curves. is positively oriented if traversing the boundary in the direction of parametrization keeps the region immediately to the left. For a simply connected region with a single outer boundary, positive orientation corresponds to counterclockwise traversal.
Statement: Let be a positively oriented, piecewise smooth, simple closed curve in the plane, enclosing a region . Let be a vector field with continuous first partial derivatives on an open set containing . Then: In vector notation, with :
Proof Strategy: It suffices to prove two independent identities and sum them:
Step 1: Prove Identity (I) for a Type I Region: Suppose , where are continuous and piecewise smooth. The boundary consists of four curves: the bottom curve (from to ), vertical segment (), the top curve (from down to ), and vertical segment ().
Evaluate the double integral on the right of (I): By the Fundamental Theorem of Calculus applied to the inner integral: Thus:
Now evaluate the line integral : On vertical segments and , is constant, so and their contributions vanish. On , runs from to ; on , runs from to : Reversing the bounds on the second integral:
Step 2: Prove Identity (II) for a Type II Region: By an identical argument on , integrating with respect to first yields . Summing (I) and (II) proves Green's Theorem for any region that is simultaneously Type I and Type II. Any standard elementary domain can be partitioned into finitely many such subregions; interior boundary line integrals cancel pairwise in opposite directions, yielding the global result.
Problem: Verify Green's Theorem for around the triangular boundary with vertices , oriented counterclockwise.
Analytical Verification:
Method 1: Double Integral Side: With and : The triangular region is bounded by and :
Method 2: Line Integral Along the Three Segments:
- (from to ): .
- (from to ): , with running from 2 to 0:
- (from to ): .
Flux-Divergence Form (2D Gauss's Theorem): Let be the outward unit normal vector to . Then the outward flux of across equals the double integral of its divergence:
Boundary Planar Area Formulas: Choosing vector fields where transforms area computation into a single boundary line integral: The symmetric formulation provides the mathematical basis for mechanical planimeters and polygonal surveyor algorithms.
Problem: Use the Green's Theorem symmetric area formula to find the area enclosed by the astroid , parametrized by , for .
Analytical Solution:
- Compute differentials:
- Form the integrand :
- Use double-angle identity: .
- Integrate over :
Thus the enclosed area is , derived directly without computing singular Cartesian bounds.
Multiply Connected Domains: If a region contains holes bounded by interior curves , the positively oriented boundary consists of the outer boundary traversed counterclockwise and all interior boundary curves traversed clockwise (so the interior of always lies to the left): where each on the right side is oriented standardly (counterclockwise).
Pathological Vortex Circulation Resolved: Consider the vortex field on an arbitrary simple closed loop enclosing the origin. Since on the annular region between and an arbitrarily small circle centered at the origin: Thus, Green's Theorem proves that the circulation around any loop enclosing the origin is topologically invariant and equals .
1. Sign Confusion in 2D Curl: The integrand in Green's theorem is , NOT . Remember the mnemonic: "Right minus Left" from the determinant of .
2. Clockwise Traversal Penalty: If a closed path is oriented clockwise (negative orientation), Green's Theorem applies with an explicit negative sign: .
3. Non-Closed Curves: Green's Theorem requires a CLOSED path. If is an open curve from to , you must close the path by adding a straight line segment from to , compute via Green's Theorem, and subtract .
4. Singularities Inside the Domain: If or fails to be at even a single point inside (e.g. division by zero at the origin), Green's Theorem cannot be applied directly to the full region. You must excise the singularity with an interior boundary circle.
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Identify and . Notice that direct line integration would be intractable due to the transcendental terms.
Compute the 2D curl: Notice that both complicated transcendental terms vanish upon differentiation.
Apply Green's Theorem over the disk : In polar coordinates: , : Since , the integral evaluates to .
The curve is open. Close the loop by defining as the line segment along the -axis from to . The composite curve forms a positively oriented boundary enclosing the upper half-disk .
By Green's Theorem: Along : , and runs from to : Since :
Surface Integrals, the Divergence Theorem & Stokes' Theorem
This capstone chapter elevates the calculus of curves and regions into three dimensions through two monumental duality principles: The Divergence Theorem (Gauss's Theorem), which equates the outward flux across a closed bounding surface to the net differential divergence throughout the enclosed volume ; and Stokes' Theorem, which equates the boundary line integral along to the net surface flux of the curl throughout the spanning surface . Together, these form the geometric and analytical backbone of classical electrodynamics, fluid mechanics, and differential geometry, unified by the generalized Stokes' formula .
This capstone chapter progresses through three instructional tiers:
- Level 1 (Foundations): Parametric surfaces , the fundamental surface normal vector , differential surface area , graph surfaces , and scalar vs. flux surface integrals.
- Level 2 (Medium / Analytical Application): The Divergence Theorem and fluid continuity, Stokes' Theorem with right-hand orientation, and surface independence (swapping an intricate curved surface for a simple planar disk sharing the same boundary).
- Level 3 (Advanced / University Rigor): Rigorous analytical proofs of both theorems, inverse-square fields and Gauss's Law with distributional point sources (Dirac delta ), and the Grand Synthesis of Vector Calculus into the exterior derivative of differential forms.
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A smooth surface is parametrized by a vector-valued function defined on a parameter domain . The tangent vectors along the grid curves are and . The fundamental normal vector is: The surface is regular if everywhere on . The differential surface area element is the area of the infinitesimal parallelogram spanned by and : For a surface defined as the graph of a function over , setting yields , giving the standard formula:
Let be an oriented surface with continuous unit normal vector field . The flux of a continuous vector field through is defined as: Notice that the scalar norm cancels algebraically, reducing flux computation directly to a standard double integral of the dot product .
Statement: Let be a solid bounded region whose boundary is a piecewise smooth, closed surface oriented with outward-pointing unit normal . If is a vector field on an open region containing , then: where .
Physical Interpretation: The outward flux across the boundary equals the total accumulation of internal sources (points where ) minus sinks (where ). In incompressible fluid flow (), net outward flux across any closed surface vanishes identically.
Proof Strategy: Decompose . It suffices to prove: and sum the analogous results for and .
Step 1: Type I Solid: Let be a Type I solid bounded between two graph surfaces (bottom surface ) and (top surface ) over a planar domain , connected by vertical lateral sides .
Step 2: Evaluate the Triple Integral: By Fubini's Theorem:
Step 3: Evaluate the Boundary Surface Flux: The boundary splits into , , and :
- On the top surface , the outward normal has positive -component: . Therefore:
- On the bottom surface , the outward normal points downward: . Therefore:
- On the vertical lateral surface , the outward normal is horizontal, so . Thus .
Summing the three boundary pieces: Permuting coordinates for Type II and Type III components and summing yields .
Statement: Let be an oriented piecewise smooth surface bounded by a simple, closed, piecewise smooth boundary curve whose orientation is compatible with the orientation of (governed by the right-hand rule: when walking along with your head pointing in the direction of the surface normal , the surface remains to your left). If is a vector field on an open region containing , then: Surface Independence Corollary: If two different oriented surfaces and share the exact same oriented boundary curve , then: This allows difficult surface integrals of curl fields to be replaced by simple flat disks or elementary planar surfaces.
Problem: Evaluate , where and is the hemisphere oriented with upward normal.
Analytical Solution:
- Identify the boundary: The boundary of the hemisphere is the circle in the -plane: oriented counterclockwise when viewed from above (by the right-hand rule with upward normal).
- Replace with the flat disk : Rather than integrating over the curved spherical hemisphere or computing the boundary line integral directly, use the flat disk having the exact same boundary . On , the upward normal is simply , and .
- Compute the curl of :
- Evaluate the dot product with normal:
- Evaluate the surface integral:
Stokes' Theorem collapses what would be an arduous spherical integral into an immediate planar multiplication.
Inverse-Square Field: Consider the classical gravitational / electrostatic vector field:
Step 1: Divergence Away from Origin: For , compute using : Summing across :
Step 2: Flux through a Sphere of Radius : On a sphere centered at the origin, the outward unit normal is , and : Thus:
Conclusion: The flux is , completely independent of . By the Divergence Theorem applied to the region between any arbitrary closed surface enclosing the origin and an inscribed sphere : In the language of distributions, , proving Gauss's Law of Electrodynamics from pure vector calculus.
1. Confusing Closed vs. Open Surfaces: The Divergence Theorem requires a CLOSED surface enclosing a 3D solid volume. If the problem asks for flux through an OPEN surface (e.g., an open bowl or cylinder without top/bottom lids), you cannot apply the Divergence Theorem directly without adding capping surfaces and subtracting their fluxes.
2. Stokes' Theorem Requires an OPEN Surface: Applying Stokes' Theorem to a closed surface (like a sphere) produces , because a closed surface has no boundary ().
3. Orientation Mismatch in Stokes' Theorem: The orientation of and the normal vector must strictly obey the right-hand rule. If you parametrize the boundary in the clockwise direction looking down an upward normal, your integral will have the wrong sign.
4. Singularities and the Divergence Theorem: If is undefined at even one point inside the solid region (e.g. in inverse-square fields), the classic Divergence Theorem is FALSE.
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
A displacement has horizontal component and vertical component . Find the squared magnitude and the area of the right triangle formed with the axes.
Step-by-step Solution:
- By the Pythagorean theorem, .
- The enclosed right-triangle area is .
Direct surface integration is prohibitive due to cross-coupling transcendental terms.
Compute the divergence: Notice that all transcendental cross-terms vanish identically.
Apply the Divergence Theorem: Convert to spherical coordinates where and over the ball of radius :
Compute : Choose the planar surface bounded by on the plane . Here . Upward normal vector: Dot product: On the plane , this is . By symmetry over the unit disk , . In polar coordinates : . Alternatively, notice that where . Velocity: . The first two terms give , integrating to . For the third term, by substitution . Total circulation = .
Practice set
Twenty-five university-rigorous problems spanning all eight chapters — from foundational partial differentiation to Stokes' Theorem and the Divergence Theorem. Try each problem before revealing the detailed worked solution.
Let defined on . Compute the first-order partial derivatives and , and evaluate and .
Basic1. Partial Derivative with respect to (treating as constant): At :
2. Partial Derivative with respect to (treating as constant): At :
Verify Clairaut's Theorem on equality of mixed partial derivatives for the function by computing and directly.
Basic1. Compute and :
2. Compute and :
Since is on , , confirming Clairaut's Theorem. ✓
Prove that the limit does not exist by using the Two-Path Test, comparing straight-line approaches with parabolic approaches.
Intermediate1. Approach along linear paths : Along (the -axis), . Along all straight lines through the origin, the limit is 0.
2. Approach along parabolic paths : Choosing yields limit ; choosing yields .
Since different approach paths yield different limiting values, the multivariable limit does not exist.
Find the directional derivative of at the point in the direction of the vector .
Basic1. Compute the gradient vector : At :
2. Normalize the direction vector :
3. Compute the directional derivative:
Let be a differentiable function, with polar coordinates transformation and . Express and in terms of and , and prove that:
Intermediate1. Apply the Multivariable Chain Rule:
2. Divide by and square both expressions:
3. Sum the squares: Cross-terms cancel identically:
A temperature distribution is given by . At the point , find: (a) the direction of steepest temperature increase, (b) the maximum rate of increase, and (c) the equation of the tangent plane to the isothermal level surface passing through .
Advanced1. Compute the gradient : Let . At , . At :
(a) Direction of steepest increase:
(b) Maximum rate of increase:
(c) Tangent plane to the isothermal surface: The level surface is . The normal to this level surface is parallel to , hence parallel to . Equation:
Find all critical points of the function .
Basic1. Set the first partial derivatives equal to zero:
2. Substitute into the second equation: Real solutions are and .
3. Determine corresponding -values:
- If , .
- If , .
Classify the critical points and of using the Second Derivative Test and Hessian matrix.
Intermediate1. Compute the second partial derivatives: The Hessian determinant discriminant is:
2. Evaluate at : Since , is a saddle point.
3. Evaluate at : Since and , is a local minimum with value .
Use the method of Lagrange multipliers to find the maximum volume of a rectangular box with faces parallel to the coordinate planes inscribed in the ellipsoid .
AdvancedLet the corner of the box in the first octant be with . The total volume is . The constraint is .
1. Set up the Lagrange multiplier system :
2. Multiply the three equations by respectively: Since , , which implies:
3. Substitute into the ellipsoid constraint:
4. Calculate the maximum volume:
Evaluate the double integral over the rectangular region .
Intermediate1. Set up the iterated integral:
2. Integrate with respect to :
3. Integrate with respect to using integration by parts:
- .
- .
Evaluate the non-elementary integral by reversing the order of integration.
Intermediate1. Sketch the region of integration : The given bounds are for . This describes the triangle bounded by the lines , , and .
2. Reverse the order to : In terms of vertical cross-sections, ranges from to , and for each fixed , ranges from up to the line :
3. Evaluate the inner integral:
4. Evaluate the outer integral: Substitute , :
Derive the Gaussian integral from first principles by evaluating the double integral in polar coordinates.
Advanced1. Express as a double integral: By Fubini's theorem on independent variables:
2. Convert to polar coordinates: Let , so and . The domain maps to and :
3. Evaluate the radial integral: Let , : Taking the limit as , this equals .
4. Conclude:
Evaluate the Cartesian triple integral over the rectangular box .
BasicSince the region is a Cartesian box and the integrand is completely separable into univariate functions , Fubini's Theorem allows complete factoring:
Compute each single-variable factor:
- .
- .
- .
Find the volume and the -coordinate of the centroid of the solid region bounded below by the -plane and above by the paraboloid , assuming uniform density .
Intermediate1. Cylindrical Coordinate Setup: The paraboloid intersects at . Bounds: , , , with .
2. Volume:
3. Moment : Let , :
4. Centroid :
Find the total mass and the -coordinate of the center of mass of the upper solid hemisphere , , whose density is proportional to the distance from the origin: for constant .
Advanced1. Coordinate Setup: In spherical coordinates: , , , .
2. Total Mass :
3. Moment with respect to -plane : Since :
4. Centroid :
Compute the moment of inertia about the -axis of a solid homogeneous right circular cone of base radius , height , and uniform total mass .
Advanced1. Cone Geometry & Density: The cone has vertex at the origin and base at , so the lateral surface is . Total volume , so density .
2. Cylindrical Integral Setup for : Bounds: , , :
3. Substitute :
Evaluate the line integral for the vector field along the circular helix for .
Basic1. Velocity vector:
2. Evaluate along the curve: With :
3. Compute dot product:
4. Integrate:
Compute the scalar line integral with respect to arc length , where is the straight line segment from the origin to .
Basic1. Parametrize the line segment:
2. Compute differential arc length :
3. Evaluate integrand along :
4. Integrate:
Prove that the 3D vector field is conservative on , construct a potential function satisfying , and evaluate along any path from to .
Intermediate1. Verify : Since is simply connected, is conservative.
2. Reconstruct the potential function:
- Integrate .
- Differentiate with respect to : .
- Differentiate with respect to : .
3. Apply the Fundamental Theorem of Line Integrals:
Use Green's Theorem to evaluate the circulation , where is the circle oriented counterclockwise.
Basic1. Identify components and compute 2D curl: With and : The non-elementary transcendental terms vanish completely.
2. Apply Green's Theorem: The region is the circular disk of radius :
Use Green's Theorem to evaluate , where is the positively oriented boundary of the region enclosed between the parabolas and .
Intermediate1. Compute the 2D curl: With and :
2. Intersections of boundary curves: and . Region is described by and .
3. Apply Green's Theorem:
Use the Green's Theorem line integral formula for area, , to derive the area enclosed by the hypocycloid (astroid) for .
Advanced1. Compute the differentials:
2. Evaluate :
3. Use trigonometric identities: Since :
4. Integrate around the closed loop:
Find the surface area of the paraboloid cap that lies below the horizontal plane .
Intermediate1. Surface Area Differential for a Graph :
2. Projection onto -plane: The intersection is a disk of radius . In polar coordinates: , , and .
3. Evaluate the surface integral: Substitute , : when ; when :
Use the Divergence Theorem to calculate the outward flux of across the closed boundary surface of the solid cylinder .
Advanced1. Compute the divergence : Notice that all cross-coupling transcendental terms vanish identically.
2. Apply the Divergence Theorem:
3. Convert to cylindrical coordinates: In , , , , , and :
Use Stokes' Theorem to evaluate , where and is the intersection of the cylinder and the plane , oriented counterclockwise when viewed from above. Verify surface independence by comparing the flux of through the flat disk and the paraboloid cap .
Advanced1. Compute :
2. Flux through flat disk (, ): On , . In polar coordinates:
3. Flux through paraboloid (): The upward normal is . Since projects onto the exact same unit disk , this yields identically .
Both surfaces give the exact same circulation , rigorously illustrating the surface independence property of Stokes' Theorem. ✓
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