Multivariable Calculus

Level · University · builds on Linear Algebra Chapters · 4 Feeds into · Differential Equations · Physics · Machine Learning

Functions of Several Variables & Partial Derivatives

A function of two variables $f(x,y)$ assigns a number to every point of a region in the plane — think of it as a surface, or a landscape with $f$ as elevation. A partial derivative measures how that surface tilts if you walk in only one coordinate direction at a time, holding the other fixed. Everything in this track builds from that one idea.

● Definition 1.1 — Function of two variables

A function $f:D\to\mathbb{R}$ with $D\subseteq\mathbb{R}^2$ assigns a real number $f(x,y)$ to each point $(x,y)\in D$ . Its graph $z=f(x,y)$ is a surface in $\mathbb{R}^3$ . Fixing $y=b$ and letting $x$ vary traces a curve on that surface — the "trace" in the direction of $x$ — and similarly for fixing $x=a$ .

● Definition 1.2 — Partial derivative

The partial derivative of $f$ with respect to $x$ at $(a,b)$ is $f_x(a,b)=\frac{\partial f}{\partial x}(a,b)=\lim_{h\to0}\frac{f(a+h,b)-f(a,b)}{h},$ and similarly $f_y(a,b)=\lim_{h\to0}\dfrac{f(a,b+h)-f(a,b)}{h}$ . In practice: to find $f_x$ , differentiate with respect to $x$ treating $y$ as a constant; for $f_y$ , treat $x$ as a constant.

Reading partial derivatives

Geometric meaning

f_x(a,b)

is the slope, in the

x

-direction, of the tangent line to the trace curve through

(a,b,f(a,b))

Second partials

f_{xx}=\partial(f_x)/\partial x

f_{xy}=\partial(f_x)/\partial y

f_{yx}=\partial(f_y)/\partial x

f_{yy}=\partial(f_y)/\partial y

The gradient

The vector

\nabla f=(f_x,f_y)

packages both partials — the subject of Chapter 2

Worked example 1.A — Computing partial derivatives

For $f(x,y)=x^2y+3xy^2-y^3$ , find $f_x(1,2)$ and $f_y(1,2)$ .

Treating $y$ as constant: $f_x=2xy+3y^2$ . Treating $x$ as constant: $f_y=x^2+6xy-3y^2$ .

$f_x(1,2)=2(1)(2)+3(2)^2=4+12=16, \qquad f_y(1,2)=1^2+6(1)(2)-3(2)^2=1+12-12=1.$

The two "mixed" second partials, $f_{xy}$ and $f_{yx}$ , are computed in opposite orders — yet for every function you'll meet in this course they come out equal. That's not a coincidence; it's a theorem, and it's what makes multivariable calculus's bookkeeping (Hessians in Chapter 3, curl and divergence beyond this track) manageable.

▲ Theorem 1.1 — Clairaut's Theorem (equality of mixed partials)

If $f_{xy}$ and $f_{yx}$ both exist near $(a,b)$ and are continuous at $(a,b)$ , then $f_{xy}(a,b)=f_{yx}(a,b).$

Proof

Fix $(a,b)$ , and for small nonzero $h,k$ consider the mixed second difference $\Delta(h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b).$

Route 1: $x$ first, then $y$ . Let $g(t)=f(t,b+k)-f(t,b)$ , so $\Delta(h,k)=g(a+h)-g(a)$ . Since $f_x$ exists, $g$ is differentiable with $g'(t)=f_x(t,b+k)-f_x(t,b)$ ; by the Mean Value Theorem, $\Delta(h,k)=h\,g'(a+\theta h)=h\big[f_x(a+\theta h,b+k)-f_x(a+\theta h,b)\big]$ for some $\theta\in(0,1)$ . Since $f_{xy}$ exists, apply the Mean Value Theorem again — this time in $y$ , to the function $s\mapsto f_x(a+\theta h,s)$ — to get $\Delta(h,k)=hk\,f_{xy}(a+\theta h,\,b+\psi k)$ for some $\psi\in(0,1)$ .

Route 2: $y$ first, then $x$ . The identical argument with the roles of $x$ and $y$ swapped — starting from $\Delta(h,k)=\big[f(a+h,b+k)-f(a,b+k)\big]-\big[f(a+h,b)-f(a,b)\big]$ — gives $\Delta(h,k)=hk\,f_{yx}(a+\theta' h,\,b+\psi' k)$ for some $\theta',\psi'\in(0,1)$ , using that $f_y$ and $f_{yx}$ exist.

Both routes compute the same quantity $\Delta(h,k)$ , so dividing by $hk\neq0$ : $f_{xy}(a+\theta h,\,b+\psi k)=f_{yx}(a+\theta' h,\,b+\psi' k)$ for every small nonzero $h,k$ . As $h,k\to0$ , both evaluation points approach $(a,b)$ , so continuity of $f_{xy}$ and $f_{yx}$ at $(a,b)$ gives $f_{xy}(a,b)=\lim_{h,k\to0}f_{xy}(a+\theta h,b+\psi k)=\lim_{h,k\to0}f_{yx}(a+\theta' h,b+\psi' k)=f_{yx}(a,b).$

Worked example 1.B — Checking Clairaut's theorem

Verify $f_{xy}=f_{yx}$ for $f(x,y)=x^3y^2+\sin(xy)$ .

$f_x=3x^2y^2+y\cos(xy)$ , so $f_{xy}=\frac{\partial}{\partial y}\big[3x^2y^2+y\cos(xy)\big]=6x^2y+\cos(xy)-xy\sin(xy).$

$f_y=2x^3y+x\cos(xy)$ , so $f_{yx}=\frac{\partial}{\partial x}\big[2x^3y+x\cos(xy)\big]=6x^2y+\cos(xy)-xy\sin(xy).$

Identical, exactly as Theorem 1.1 guarantees — both $f_{xy}$ and $f_{yx}$ are continuous everywhere here (polynomials and $\sin,\cos$ of polynomials), so the theorem applies at every point.

📝 Chapter Quiz

Questions Level

The Chain Rule & Directional Derivatives

Chapter 1 measured how $f$ changes along the two coordinate directions. This chapter does two things with that: differentiates $f$ along a path that moves through both $x$ and $y$ at once (the chain rule), and measures the rate of change along any direction at all (the directional derivative) — which turns out to be answered completely by a single vector, the gradient.

● Definition 2.1 — Multivariable chain rule

If $z=f(x,y)$ is differentiable and $x=x(t)$ , $y=y(t)$ are differentiable functions of $t$ , then $z(t)=f(x(t),y(t))$ is differentiable with $\frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}=f_x\,x'(t)+f_y\,y'(t).$ Each term accounts for the change coming from one input variable, holding the other's contribution separate.

Worked example 2.A — Applying the chain rule

For $z=x^2y$ with $x=t^2$ , $y=t^3$ , find $dz/dt$ two ways.

Direct substitution: $z=(t^2)^2(t^3)=t^7$ , so $dz/dt=7t^6$ .

Chain rule: $z_x=2xy$ , $z_y=x^2$ , $x'=2t$ , $y'=3t^2$ , so $\frac{dz}{dt}=(2xy)(2t)+(x^2)(3t^2)=2(t^2)(t^3)(2t)+(t^2)^2(3t^2)=4t^6+3t^6=7t^6.$

Both routes agree.

● Definition 2.2 — Gradient and directional derivative

The gradient of $f$ at $(a,b)$ is the vector $\nabla f(a,b)=\big(f_x(a,b),\,f_y(a,b)\big)$ . For a unit vector $\mathbf{u}=(u_1,u_2)$ , the directional derivative of $f$ at $(a,b)$ in the direction $\mathbf{u}$ is $D_{\mathbf{u}}f(a,b)=\nabla f(a,b)\cdot\mathbf{u}=f_x(a,b)u_1+f_y(a,b)u_2,$ the instantaneous rate of change of $f$ as you move from $(a,b)$ at unit speed in the direction $\mathbf{u}$ . Taking $\mathbf{u}=(1,0)$ or $(0,1)$ recovers $f_x$ and $f_y$ as the special cases they always were.

Why the gradient

It's a vector

\nabla f

lives in

\mathbb{R}^2

, so everything from Linear Algebra Chapter 1 — dot products, norms, Cauchy–Schwarz — applies to it directly

It answers every direction at once

One vector

\nabla f(a,b)

determines

D_{\mathbf{u}}f(a,b)

for every unit

\mathbf{u}

, via a single dot product

Which direction makes $f$ increase fastest? The dot-product formula for $D_{\mathbf{u}}f$ makes this a question Linear Algebra already answered.

▲ Theorem 2.1 — The gradient points in the direction of steepest ascent

If $\nabla f(a,b)\neq\mathbf{0}$ , then among all unit vectors $\mathbf{u}$ , the directional derivative $D_{\mathbf{u}}f(a,b)$ is maximized exactly when $\mathbf{u}=\dfrac{\nabla f(a,b)}{\lVert\nabla f(a,b)\rVert}$ , and that maximum value is $\lVert\nabla f(a,b)\rVert$ .

Proof

Write $\mathbf{g}=\nabla f(a,b)$ . By the Cauchy–Schwarz inequality (Linear Algebra, Theorem 1.1), for any unit vector $\mathbf{u}$ (so $\lVert\mathbf{u}\rVert=1$ ): $D_{\mathbf{u}}f(a,b)=\mathbf{g}\cdot\mathbf{u}\le|\mathbf{g}\cdot\mathbf{u}|\le\lVert\mathbf{g}\rVert\,\lVert\mathbf{u}\rVert=\lVert\mathbf{g}\rVert.$ So no direction can push $D_{\mathbf{u}}f(a,b)$ past $\lVert\mathbf{g}\rVert$ .

That bound is actually attained: since $\mathbf{g}\neq\mathbf{0}$ , the unit vector $\mathbf{u}^*=\mathbf{g}/\lVert\mathbf{g}\rVert$ is well-defined, and $D_{\mathbf{u}^*}f(a,b)=\mathbf{g}\cdot\frac{\mathbf{g}}{\lVert\mathbf{g}\rVert}=\frac{\mathbf{g}\cdot\mathbf{g}}{\lVert\mathbf{g}\rVert}=\frac{\lVert\mathbf{g}\rVert^2}{\lVert\mathbf{g}\rVert}=\lVert\mathbf{g}\rVert.$ So $\lVert\mathbf{g}\rVert$ is both an upper bound for $D_{\mathbf{u}}f(a,b)$ over all unit $\mathbf{u}$ , and a value it actually reaches at $\mathbf{u}=\mathbf{u}^*$ — making it the maximum, attained exactly there.

Worked example 2.B — Directional derivative and steepest ascent

For $f(x,y)=x^2+xy$ at $(1,2)$ , find the directional derivative toward $\mathbf{v}=(3,4)$ , and the maximum possible rate of increase in any direction.

$\nabla f=(2x+y,\,x)$ , so $\nabla f(1,2)=(4,1)$ . The unit vector toward $\mathbf{v}$ is $\mathbf{u}=\mathbf{v}/\lVert\mathbf{v}\rVert=(3,4)/5=(3/5,4/5)$ (since $\lVert\mathbf{v}\rVert=\sqrt{3^2+4^2}=5$ ). So $D_{\mathbf{u}}f(1,2)=(4)(3/5)+(1)(4/5)=\frac{12+4}{5}=\frac{16}{5}=3.2.$

By Theorem 2.1, the maximum rate of increase in any direction is $\lVert\nabla f(1,2)\rVert=\sqrt{4^2+1^2}=\sqrt{17}\approx4.12$ — larger than $3.2$ , since $\mathbf{v}$ doesn't point exactly along $\nabla f(1,2)$ .

📝 Chapter Quiz

Questions Level

Optimization: Critical Points & Lagrange Multipliers

Finding a maximum or minimum of $f(x,y)$ starts the same way it did in single-variable calculus — at a smooth extremum, the derivative vanishes. The difference is that now "the derivative" is a vector, and once it vanishes, you need a second test to tell a peak from a valley from a saddle. That test turns out to be a $2\times2$ determinant, straight out of Linear Algebra Chapter 4.

● Definition 3.1 — Critical point

$(a,b)$ is a critical point of $f$ if $\nabla f(a,b)=\mathbf{0}$ — that is, $f_x(a,b)=f_y(a,b)=0$ — or if a partial derivative fails to exist there. Every local max or min where $f$ is differentiable occurs at a critical point (if some direction had $D_{\mathbf{u}}f\neq0$ , moving that way would increase or decrease $f$ , so the point couldn't be extremal).

● Definition 3.2 — Hessian and discriminant

At a point where the second partials exist, the Hessian matrix is $H=\begin{pmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{pmatrix},$ and (using $f_{xy}=f_{yx}$ from Clairaut's theorem) its discriminant is $D=\det(H)=f_{xx}f_{yy}-f_{xy}^2.$

The sign of $D$ , together with the sign of $f_{xx}$ , classifies every non-degenerate critical point.

▲ Theorem 3.1 — The Second Derivative Test

Let $(a,b)$ be a critical point of $f$ , with $f$ 's second partials continuous near $(a,b)$ , and let $D=D(a,b)$ as in Definition 3.2.

If $D>0$ and $f_{xx}(a,b)>0$ : $f$ has a local minimum at $(a,b)$ .
If $D>0$ and $f_{xx}(a,b)<0$ : $f$ has a local maximum at $(a,b)$ .
If $D<0$ : $(a,b)$ is a saddle point (neither a local max nor min).
If $D=0$ : the test is inconclusive.

Proof

Write $h,k$ for small displacements and drop the argument $(a,b)$ from $f_{xx},f_{xy},f_{yy}$ . Since $(a,b)$ is critical, the first-order terms of $f$ 's Taylor expansion vanish, and — provided $D\neq0$ — the quadratic term dominates the remainder for $(h,k)$ small, so the sign of $f(a+h,b+k)-f(a,b)$ matches the sign of the quadratic form $Q(h,k)=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2.$

Case $f_{xx}\neq0$ . Completing the square in $h$ : $Q(h,k)=f_{xx}\Big(h+\frac{f_{xy}}{f_{xx}}k\Big)^{2}+\Big(f_{yy}-\frac{f_{xy}^2}{f_{xx}}\Big)k^2=f_{xx}\Big(h+\frac{f_{xy}}{f_{xx}}k\Big)^{2}+\frac{D}{f_{xx}}k^2,$ using $f_{yy}-f_{xy}^2/f_{xx}=(f_{xx}f_{yy}-f_{xy}^2)/f_{xx}=D/f_{xx}$ .

If $D>0$ and $f_{xx}>0$ , both $f_{xx}$ and $D/f_{xx}$ are positive, so $Q(h,k)$ is a sum of two nonnegative terms, zero only when both squared terms vanish — which (since $D/f_{xx}\neq0$ ) forces $k=0$ , then $h=0$ . So $Q(h,k)>0$ for every $(h,k)\neq(0,0)$ : a local minimum. If $D>0$ and $f_{xx}<0$ , both coefficients are negative by the same reasoning, so $Q(h,k)<0$ away from the origin: a local maximum. If $D<0$ , then $f_{xx}$ and $D/f_{xx}$ have opposite signs: along $k=0$ , $Q=f_{xx}h^2$ has the sign of $f_{xx}$ ; along $h=-(f_{xy}/f_{xx})k$ , $Q=(D/f_{xx})k^2$ has the opposite sign. $Q$ takes both signs arbitrarily close to $(0,0)$ : a saddle point.

Case $f_{xx}=0$ . Then $D=-f_{xy}^2\le0$ automatically, so only the $D\le0$ rows of the theorem are reachable here, and only $D<0$ (i.e. $f_{xy}\neq0$ ) is claimed. Here $Q(h,k)=2f_{xy}hk+f_{yy}k^2$ . Taking $(h,k)=(1,\varepsilon)$ for small $\varepsilon\neq0$ gives $Q\approx2f_{xy}\varepsilon$ , whose sign flips with the sign of $\varepsilon$ ; so $Q$ again takes both signs arbitrarily close to the origin — a saddle point, matching the theorem.

Worked example 3.A — Classifying critical points

Classify the critical points of $f(x,y)=x^3-3x+y^2$ .

$f_x=3x^2-3=0\Rightarrow x=\pm1$ ; $f_y=2y=0\Rightarrow y=0$ . Critical points: $(1,0)$ and $(-1,0)$ .

$f_{xx}=6x$ , $f_{yy}=2$ , $f_{xy}=0$ , so $D=12x$ .

At $(1,0)$ : $D=12>0$ and $f_{xx}=6>0$ — local minimum, value $f(1,0)=1-3=-2$ .

At $(-1,0)$ : $D=-12<0$ — saddle point.

The second derivative test optimizes $f$ freely over the whole plane. Often, though, you need the best value of $f$ only among points satisfying some constraint $g(x,y)=c$ — walking along a fixed curve rather than roaming the plane.

▲ Theorem 3.2 — Lagrange Multipliers

Suppose $f$ has a local extremum, subject to the constraint $g(x,y)=c$ , at a point $(a,b)$ where $\nabla g(a,b)\neq\mathbf{0}$ . Then there exists a scalar $\lambda$ with $\nabla f(a,b)=\lambda\,\nabla g(a,b).$

Proof

Since $\nabla g(a,b)\neq\mathbf{0}$ , the level curve $g(x,y)=c$ has a smooth parametrization $\mathbf{r}(t)=(x(t),y(t))$ near $(a,b)$ , with $\mathbf{r}(t_0)=(a,b)$ and $\mathbf{r}'(t_0)\neq\mathbf{0}$ tangent to the curve.

Because $g(\mathbf{r}(t))=c$ is constant along this curve, the chain rule (Definition 2.1) gives $\nabla g(a,b)\cdot\mathbf{r}'(t_0)=0$ : the gradient $\nabla g(a,b)$ is orthogonal to $\mathbf{r}'(t_0)$ .

Since $(a,b)$ is a local extremum of $f$ restricted to the constraint, $h(t)=f(\mathbf{r}(t))$ has a local extremum at $t=t_0$ , so $h'(t_0)=0$ . By the chain rule again, $h'(t_0)=\nabla f(a,b)\cdot\mathbf{r}'(t_0)=0$ : the gradient $\nabla f(a,b)$ is also orthogonal to $\mathbf{r}'(t_0)$ .

So both $\nabla f(a,b)$ and $\nabla g(a,b)$ are orthogonal to the same nonzero vector $\mathbf{r}'(t_0)$ in $\mathbb{R}^2$ . In the plane, the set of vectors orthogonal to a given nonzero vector is a single line through the origin (a 1-dimensional subspace) — so any two vectors orthogonal to $\mathbf{r}'(t_0)$ are scalar multiples of each other. Hence $\nabla f(a,b)=\lambda\,\nabla g(a,b)$ for some scalar $\lambda$ (taking $\lambda=0$ if $\nabla f(a,b)=\mathbf{0}$ ).

Worked example 3.B — Using a Lagrange multiplier

Maximize $f(x,y)=xy$ subject to $x+y=10$ .

Here $g(x,y)=x+y$ , so $\nabla g=(1,1)$ — never $\mathbf{0}$ , so Theorem 3.2 applies at any extremum. $\nabla f=(y,x)$ . Setting $\nabla f=\lambda\nabla g$ : $y=\lambda,\qquad x=\lambda\quad\Longrightarrow\quad x=y.$ Combined with $x+y=10$ : $x=y=5$ , so $f(5,5)=25$ is the constrained maximum (checking nearby constrained points, e.g. $(9,1)$ gives $f=9$ and $(6,4)$ gives $f=24$ , both less than $25$ , consistent with a maximum here).

📝 Chapter Quiz

Questions Level

Double Integrals

A single integral $\int_a^b f(x)\,dx$ sums up $f$ over an interval; a double integral $\iint_R f(x,y)\,dA$ sums $f$ over a whole region $R$ of the plane, and (for $f\ge0$ ) gives the volume under the surface $z=f(x,y)$ . The one new practical question is how to actually compute one — and the answer is to reduce it to two ordinary, single-variable integrals, done one after the other.

● Definition 4.1 — Double integral over a rectangle

For $f$ continuous on $R=[a,b]\times[c,d]$ , partition $R$ into an $m\times n$ grid of sub-rectangles $R_{ij}$ , each of area $\Delta A=\Delta x\,\Delta y$ , and pick a sample point $(x_i^*,y_j^*)\in R_{ij}$ . The double integral is the limit of the Riemann sum as the grid is refined: $\iint_R f\,dA=\lim_{m,n\to\infty}\sum_{i=1}^{m}\sum_{j=1}^{n}f(x_i^*,y_j^*)\,\Delta A.$

● Definition 4.2 — Iterated integral

An iterated integral computes one variable at a time: $\int_a^b\!\!\int_c^d f(x,y)\,dy\,dx$ means: first integrate $f(x,y)$ with respect to $y$ (treating $x$ as a constant) from $c$ to $d$ , producing a function of $x$ alone, then integrate that result with respect to $x$ from $a$ to $b$ .

The double integral (Definition 4.1) and the iterated integral (Definition 4.2) are computed completely differently — one is a genuine two-dimensional limit, the other is two ordinary one-dimensional integrals in sequence. The theorem that makes double integrals practical to compute is that, for continuous $f$ , they always agree.

▲ Theorem 4.1 — Fubini's Theorem

If $f$ is continuous on $R=[a,b]\times[c,d]$ , then $\iint_R f\,dA=\int_a^b\!\!\int_c^d f(x,y)\,dy\,dx=\int_c^d\!\!\int_a^b f(x,y)\,dx\,dy.$ This holds for every continuous $f$ on a rectangle — the fully general proof (which needs the uniform continuity of $f$ on the compact rectangle $R$ ) belongs to a real analysis course. Below we prove it in full for the case that actually drives how you'll split integrals apart in practice: a separable integrand $f(x,y)=g(x)h(y)$ .

Proof (separable case

f(x,y)=g(x)h(y)

)

Fix any partition of $R$ into sub-rectangles $R_{ij}$ with sample points $(x_i^*,y_j^*)$ , as in Definition 4.1. Because $f(x_i^*,y_j^*)=g(x_i^*)h(y_j^*)$ , the double Riemann sum factors exactly — for every partition, not just in a limit — by distributivity: $\sum_{i=1}^{m}\sum_{j=1}^{n}f(x_i^*,y_j^*)\,\Delta x\,\Delta y=\sum_{i=1}^{m}\sum_{j=1}^{n}g(x_i^*)h(y_j^*)\,\Delta x\,\Delta y=\Big(\sum_{i=1}^{m}g(x_i^*)\Delta x\Big)\Big(\sum_{j=1}^{n}h(y_j^*)\Delta y\Big).$

As the grid is refined ( $m,n\to\infty$ ), the two factors on the right are ordinary single-variable Riemann sums, converging (since $g,h$ are continuous, hence Riemann integrable) to $\int_a^b g(x)\,dx$ and $\int_c^d h(y)\,dy$ respectively. Since both factors converge, their product converges to the product of the limits: $\iint_R f\,dA=\Big(\int_a^b g(x)\,dx\Big)\Big(\int_c^d h(y)\,dy\Big).$

This is exactly the value of either iterated integral: $\int_a^b\!\int_c^d g(x)h(y)\,dy\,dx=\int_a^b g(x)\Big[\int_c^d h(y)\,dy\Big]dx=\Big(\int_c^d h(y)\,dy\Big)\int_a^b g(x)\,dx$ , since $\int_c^d h(y)\,dy$ is a constant with respect to $x$ and pulls out of the outer integral — and symmetrically for the other order. All three quantities are equal.

Worked example 4.A — Order of integration doesn't matter

Compute $\iint_R(x^2+y)\,dA$ for $R=[0,2]\times[0,1]$ , both orders.

$\int_0^2\!\!\int_0^1(x^2+y)\,dy\,dx=\int_0^2\Big[x^2y+\tfrac{y^2}{2}\Big]_0^1dx=\int_0^2\Big(x^2+\tfrac12\Big)dx=\Big[\tfrac{x^3}{3}+\tfrac{x}{2}\Big]_0^2=\tfrac83+1=\tfrac{11}{3}.$

$\int_0^1\!\!\int_0^2(x^2+y)\,dx\,dy=\int_0^1\Big[\tfrac{x^3}{3}+xy\Big]_0^2dy=\int_0^1\Big(\tfrac83+2y\Big)dy=\Big[\tfrac83y+y^2\Big]_0^1=\tfrac83+1=\tfrac{11}{3}.$

Same value either way, as Theorem 4.1 guarantees — $x^2+y$ isn't separable, but it's still continuous on $R$ , so the general (unproved-here) case of Fubini applies.

Polar coordinates:

dA=r\,dr\,d\theta

When a region is naturally circular, switch to polar coordinates $x=r\cos\theta$ , $y=r\sin\theta$ . A small polar "rectangle" spanning $[r,r+\Delta r]\times[\theta,\theta+\Delta\theta]$ has one pair of sides of length $\Delta r$ (radial) and the other pair of approximate length $r\,\Delta\theta$ (arc length at radius $r$ ), so its area is approximately $r\,\Delta r\,\Delta\theta$ rather than $\Delta r\,\Delta\theta$ — the extra factor of $r$ is exactly why $dA=r\,dr\,d\theta$ , not $dr\,d\theta$ , and why $x^2+y^2$ becomes the clean $r^2$ .

Worked example 4.B — A double integral in polar coordinates

Compute $\iint_D(x^2+y^2)\,dA$ where $D$ is the disk $x^2+y^2\le4$ .

In polar coordinates, $D$ is $0\le r\le2$ , $0\le\theta\le2\pi$ , and $x^2+y^2=r^2$ , $dA=r\,dr\,d\theta$ : $\iint_D(x^2+y^2)\,dA=\int_0^{2\pi}\!\!\int_0^2 r^2\cdot r\,dr\,d\theta=\int_0^{2\pi}\!\!\Big[\frac{r^4}{4}\Big]_0^2 d\theta=\int_0^{2\pi}4\,d\theta=8\pi.$

The integrand is separable in polar form ( $r^3$ times $1$ ), which is exactly why the $r$ - and $\theta$ -integrals split apart so cleanly — the same factoring idea proved in Theorem 4.1.

📝 Chapter Quiz

Questions Level

Practice set

Sixteen free-response problems spanning all four chapters — four per chapter, Basic to Advanced. Try each one, then reveal the worked solution.

Find $f_x(2,1)$ for $f(x,y)=3x^2y-y^3$ .

Basic

f_x=6xy,\qquad f_x(2,1)=6(2)(1)=12.

Find $f_y(1,3)$ for $f(x,y)=x^2+y^2$ .

Basic

f_y=2y,\qquad f_y(1,3)=2(3)=6.

Find $f_{xy}$ for $f(x,y)=e^{xy}$ .

Intermediate

$f_x=ye^{xy}$ , so $f_{xy}=\frac{\partial}{\partial y}\big[ye^{xy}\big]=e^{xy}+xye^{xy}=(1+xy)e^{xy}.$

Verify Clairaut's theorem for $f(x,y)=x^2\sin y+y^2\cos x$ by computing $f_{xy}$ and $f_{yx}$ directly.

Advanced

$f_x=2x\sin y-y^2\sin x$ , so $f_{xy}=2x\cos y-2y\sin x$ . $f_y=x^2\cos y+2y\cos x$ , so $f_{yx}=2x\cos y-2y\sin x$ . Identical, as Theorem 1.1 guarantees.

Find $\nabla f(2,1)$ for $f(x,y)=x^2y$ .

Basic

\nabla f=(2xy,\,x^2),\qquad \nabla f(2,1)=(4,4).

For $z=x^2+y^2$ with $x=t$ , $y=t^2$ , find $dz/dt$ at $t=1$ , using the chain rule.

Intermediate

$z_x=2x$ , $z_y=2y$ , $x'=1$ , $y'=2t$ . At $t=1$ : $x=1,y=1$ , so $\frac{dz}{dt}=2x(1)+2y(2t)=2(1)+2(1)(2)=6.$ (Check: $z=t^2+t^4$ , $dz/dt=2t+4t^3$ , at $t=1$ : $2+4=6$ .)

Find the directional derivative of $f(x,y)=x^2-y^2$ at $(2,-1)$ in the direction of $\mathbf{v}=(1,1)$ .

Intermediate

$\nabla f=(2x,-2y)$ , so $\nabla f(2,-1)=(4,2)$ . Unit vector $\mathbf{u}=(1,1)/\sqrt2$ . $D_{\mathbf{u}}f=\frac{4+2}{\sqrt2}=\frac{6}{\sqrt2}=3\sqrt2.$

Find the direction of steepest ascent and the maximum rate of increase of $f(x,y)=\ln(x^2+y^2)$ at $(3,4)$ .

Advanced

$\nabla f=\Big(\dfrac{2x}{x^2+y^2},\dfrac{2y}{x^2+y^2}\Big)$ . At $(3,4)$ , $x^2+y^2=25$ , so $\nabla f(3,4)=(6/25,8/25)$ , with $\lVert\nabla f(3,4)\rVert=\sqrt{(6/25)^2+(8/25)^2}=10/25=2/5$ . By Theorem 2.1: steepest ascent is in the direction $\nabla f/\lVert\nabla f\rVert=(3/5,4/5)$ , at maximum rate $2/5$ .

Find the critical point of $f(x,y)=x^2+y^2-4x+2y$ .

Basic

f_x=2x-4=0\Rightarrow x=2,\qquad f_y=2y+2=0\Rightarrow y=-1.

Critical point:

(2,-1)

M10

Classify the critical point of $f(x,y)=x^2+y^2-4x+2y$ found in M9, using the second derivative test.

Intermediate

$f_{xx}=2$ , $f_{yy}=2$ , $f_{xy}=0$ , so $D=f_{xx}f_{yy}-f_{xy}^2=4>0$ , and $f_{xx}=2>0$ . By Theorem 3.1, $(2,-1)$ is a local minimum.

M11

Find and classify the critical point(s) of $f(x,y)=y^2-x^2$ .

Intermediate

$f_x=-2x=0$ , $f_y=2y=0$ give the single critical point $(0,0)$ . $f_{xx}=-2$ , $f_{yy}=2$ , $f_{xy}=0$ , so $D=-4<0$ . By Theorem 3.1, $(0,0)$ is a saddle point.

M12

Use a Lagrange multiplier to minimize $f(x,y)=x^2+y^2$ subject to $x+2y=5$ .

Advanced

$g(x,y)=x+2y$ , $\nabla g=(1,2)\neq\mathbf{0}$ , so Theorem 3.2 applies. $\nabla f=(2x,2y)=\lambda(1,2)$ gives $2x=\lambda$ , $2y=2\lambda$ , so $y=2x$ . With $x+2y=5$ : $x+4x=5\Rightarrow x=1,\,y=2$ . Minimum value: $f(1,2)=1+4=5$ .

M13

Compute $\iint_R xy\,dA$ for $R=[0,2]\times[0,1]$ .

Basic

Separable: $\iint_R xy\,dA=\Big(\int_0^2x\,dx\Big)\Big(\int_0^1y\,dy\Big)=(2)\Big(\tfrac12\Big)=1.$

M14

Compute $\iint_R(2x+3y)\,dA$ for $R=[0,1]\times[0,2]$ .

Intermediate

$\int_0^1\!\!\int_0^2(2x+3y)\,dy\,dx=\int_0^1\big[2xy+\tfrac32y^2\big]_0^2dx=\int_0^1(4x+6)\,dx=\big[2x^2+6x\big]_0^1=8.$

M15

Compute $\iint_D e^{-(x^2+y^2)}\,dA$ where $D$ is the unit disk $x^2+y^2\le1$ , using polar coordinates.

Advanced

$\iint_D e^{-(x^2+y^2)}\,dA=\int_0^{2\pi}\!\!\int_0^1e^{-r^2}r\,dr\,d\theta.$ With $u=r^2$ , $du=2r\,dr$ : $\int_0^1e^{-r^2}r\,dr=\tfrac12(1-e^{-1})$ . So the integral is $2\pi\cdot\tfrac12(1-e^{-1})=\pi\big(1-\tfrac1e\big)\approx1.986$ .

M16

Find the area enclosed by the polar curve $r=2\cos\theta$ (a circle) using a double integral in polar coordinates.

Advanced

The curve is traced once for $\theta\in[-\pi/2,\pi/2]$ . $\text{Area}=\int_{-\pi/2}^{\pi/2}\!\!\int_0^{2\cos\theta}r\,dr\,d\theta=\int_{-\pi/2}^{\pi/2}\Big[\tfrac{r^2}{2}\Big]_0^{2\cos\theta}d\theta=\int_{-\pi/2}^{\pi/2}2\cos^2\theta\,d\theta=\pi.$ Matches the known geometry: $r=2\cos\theta$ is a circle of radius $1$ (centered at $(1,0)$ ), area $\pi(1)^2=\pi$ .

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Multivariable Calculus

Functions of Several Variables & Partial Derivatives

Reading partial derivatives

The Chain Rule & Directional Derivatives

Why the gradient

Optimization: Critical Points & Lagrange Multipliers

Double Integrals

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