[CALCULUS] Change of variables and polar coordinates

Use Jacobians to transform double integrals into coordinates that fit the region or the integrand.

This page changes the coordinates of a double integral to fit the region or the integrand. When the Cartesian slices from regions and order of integration are awkward, another coordinate system may be simpler. A disk, for example, has square-root bounds in Cartesian coordinates but constant bounds in polar coordinates.

The new coordinates simplify the description and the Jacobian corrects the area element for how small pieces stretch or shrink. The same tools evaluate weighted integrals in mass, centroids and probability and the series integral in Euler's Basel sum.

The Jacobian records area scaling

Suppose the old coordinates are given in terms of new ones by

(x,y)=g(u,v)=\bigl(x(u,v),y(u,v)\bigr).

[DEFINITION]

Jacobian matrix

The Jacobian matrix of $g$ is

J_g(u,v)= \begin{pmatrix} \dfrac{\partial x}{\partial u}& \dfrac{\partial x}{\partial v}\\[8pt] \dfrac{\partial y}{\partial u}& \dfrac{\partial y}{\partial v} \end{pmatrix}.

[1]:Jacobian determinant

[DEFINITION]

Jacobian determinant

The Jacobian determinant is

\det J_g(u,v) =\frac{\partial(x,y)}{\partial(u,v)}.

Its absolute value is the local area-scaling factor:

dA=dx\,dy =\left|\det J_g(u,v)\right|\,du\,dv.

The determinant is negative when the map reverses orientation. Area is non-negative, so the change-of-variables formula uses its absolute value.

To see why it measures area, take small increments $\Delta u,\Delta v$ :

g(u+\Delta u,v+\Delta v)-g(u,v) \approx J_g(u,v) \begin{pmatrix}\Delta u\\ \Delta v\end{pmatrix}.

The linear map $J_g(u,v)$ sends a small coordinate rectangle of area $\Delta u\,\Delta v$ to a parallelogram of area approximately $|\det J_g(u,v)|\Delta u\,\Delta v$ . Taking the limit gives the area element $dA=|\det J_g(u,v)|\,du\,dv$ .

[THEOREM]

Change of variables

Let $\Omega$ be a region in $(x,y)$ -coordinates and let $\Phi$ be a region in $(u,v)$ -coordinates. Assume that $f$ is integrable on $\Omega$ , that $g$ is $C^1$ on $\Phi$ , that $\Phi$ can be split into finitely many pieces on which $g$ is one-to-one with non-overlapping interiors and that the Jacobian is nonzero away from boundary or other exceptional sets of area zero. The images of those pieces must cover $\Omega$ once, apart from boundaries of area zero; otherwise the transformed integral can omit or count part of the region more than once. If a map covers the target with multiplicity, split the parameter region into one-to-one pieces (or account for that multiplicity) before applying the formula. Then

\iint_\Omega f(x,y)\,dx\,dy = \iint_\Phi f\bigl(x(u,v),y(u,v)\bigr) \left|\det J_g(u,v)\right|\,du\,dv.

Applying the theorem involves two separate tasks: rewriting the integrand in terms of $u,v$ and describing the old region in the new coordinates. The second task is geometric and cannot be skipped.

[1]:Jacobian determinant
[2]:Euler's Basel sum
[3]:Choose and apply a coordinate change
[4]:Polar area element

[METHOD]

Choose and apply a coordinate change

To change variables in a double integral:

choose new coordinates suggested by the boundaries or by a repeated expression in the integrand;
solve for $x$ and $y$ in terms of the new coordinates when needed;
transform every boundary of the region;
rewrite the integrand;
calculate the absolute Jacobian factor;
integrate over the transformed region.

Good coordinates simplify the region, the integrand or both. A substitution that simplifies the formula but complicates the region may not help.

Polar coordinates

[THEOREM]

Polar area element

For $r\ge0$ , the polar-coordinate change of variables is

x=r\cos\theta,\qquad y=r\sin\theta,

and its Jacobian satisfies

\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| =r.

Consequently,

dA=r\,dr\,d\theta,

and

\iint_\Omega f(x,y)\,dA =\iint_\Phi f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The factor $r$ is also visible geometrically. A small polar rectangle has radial side approximately $dr$ and angular side approximately $r\,d\theta$ , so its area is approximately $r\,dr\,d\theta$ . The same $(r,\theta)$ pair parametrises complex numbers in polar form, with $x+iy=re^{i\theta}$ .

A usual polar description takes $r\ge0$ and an angular interval of length at most $2\pi$ . Each point then has one representation, except on a boundary ray and at the origin. Those exceptional sets have area zero, so the change-of-variables theorem still applies.

The radial limits are read along each ray from the boundary of the region; they need not be constant.

[EXAMPLE]

Area of a quarter disk

The first-quadrant portion of the unit disk is

0\le r\le1,\qquad 0\le\theta\le\frac{\pi}{2}.

Its area is

\begin{aligned} \operatorname{Area}(\Omega) &=\int_0^{\pi/2}\int_0^1r\,dr\,d\theta\\ &=\int_0^{\pi/2}\frac12\,d\theta =\frac{\pi}{4}. \end{aligned}

[routine]

[1]:Polar area element

[EXAMPLE]

A triangle in polar coordinates

Evaluate

I=\int_0^1\int_0^x \left(1+\sqrt{x^2+y^2}\right)\,dy\,dx.

The region is the first-quadrant triangle bounded by $x=1$ and $y=x$ . In polar coordinates,

0\le\theta\le\frac{\pi}{4}, \qquad 0\le r\le\sec\theta,

because $x=r\cos\theta\le1$ . Also, $\sqrt{x^2+y^2}=r$ . Therefore

\begin{aligned} I &=\int_0^{\pi/4}\int_0^{\sec\theta}(1+r)r\,dr\,d\theta\\ &=\int_0^{\pi/4} \left[\frac{r^2}{2}+\frac{r^3}{3}\right]_{0}^{\sec\theta} d\theta\\ &=\int_0^{\pi/4} \left(\frac{\sec^2\theta}{2} +\frac{\sec^3\theta}{3}\right)d\theta\\ &=\frac12+\frac{\sqrt2}{6} +\frac16\ln(1+\sqrt2). \end{aligned}

[exam]

[1]:Polar area element

[EXAMPLE]

Area enclosed by a spiral

The region swept between

r=\theta,\qquad0\le\theta\le2\pi,

and the $x$ -axis has

0\le\theta\le2\pi,\qquad0\le r\le\theta.

Thus

\begin{aligned} \operatorname{Area}(\Omega) &=\int_0^{2\pi}\int_0^\theta r\,dr\,d\theta\\ &=\int_0^{2\pi}\frac{\theta^2}{2}\,d\theta =\frac{4\pi^3}{3}. \end{aligned}

The familiar polar area formula

\operatorname{Area}(\Omega) =\frac12\int r(\theta)^2\,d\theta

is the result of performing the radial integral first.

[routine]

[1]:Polar area element

[EXAMPLE]

The Gaussian integral

Let $\Omega$ be the first quadrant and consider

I=\iint_\Omega e^{-(x^2+y^2)}\,dA.

The region is unbounded, so apply the change of variables on bounded quarter-disks and then let their radius tend to infinity. The integrand is nonnegative, so monotone convergence justifies this limit. Polar coordinates give

\begin{aligned} I &=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta\\ &=\int_0^{\pi/2}\frac12\,d\theta =\frac{\pi}{4}. \end{aligned}

In Cartesian coordinates the integrand is a product $e^{-x^2}e^{-y^2}$ . Because it is nonnegative, Tonelli's theorem (or monotone limits over bounded rectangles) justifies separating the two improper integrals:

I =\left(\int_0^\infty e^{-x^2}\,dx\right)^2.

Hence

\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt\pi}{2}.

For

G=\int_{-\infty}^{\infty}e^{-x^2/2}\,dx,

symmetry and $u=x/\sqrt2$ give

\begin{aligned} G &=2\int_0^\infty e^{-x^2/2}\,dx\\ &=2\sqrt2\int_0^\infty e^{-u^2}\,du =\sqrt{2\pi}. \end{aligned}

[challenge]

[1]:Polar area element

Other coordinate changes

[EXAMPLE]

A linear change of variables

Evaluate

I=\int_0^3\int_{-x}^{x}(2-x-y)^2\,dy\,dx

using

u=x+y,\qquad v=x-y.

The inverse transformation is

x=\frac{u+v}{2}, \qquad y=\frac{u-v}{2},

and

\left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left| \det \begin{pmatrix} \frac12&\frac12\\[2pt] \frac12&-\frac12 \end{pmatrix} \right| =\frac12.

The original region is bounded by $y=-x$ , $y=x$ and $x=3$ . These become

y=-x\Longrightarrow u=0,\qquad y=x\Longrightarrow v=0,\qquad x=3\Longrightarrow u+v=6.

Thus

0\le u\le6,\qquad0\le v\le6-u.

Since $x+y=u$ , the integrand is $(2-u)^2$ . Therefore

\begin{aligned} I &=\int_0^6\int_0^{6-u}(2-u)^2\frac12\,dv\,du\\ &=\frac12\int_0^6(2-u)^2(6-u)\,du =18. \end{aligned}

These coordinates suit the problem because $x+y$ occurs in the integrand and the region becomes a triangle with two sides on the axes.

[exam]

[1]:Choose and apply a coordinate change

[EXAMPLE]

A nonlinear change of variables

Evaluate

\iint_\Omega(4x^2-2y^2)\,dA,

where

\Omega=\{(x,y):1\le xy\le4,\ x\le y\le4x\}.

Use

u=xy,\qquad v=\frac yx.

On $\Omega$ , the inequalities force $x>0$ : if $x<0$ , then $4x<x$ and the condition $x\le y\le4x$ is impossible. The map therefore has the positive inverse branch

x=\sqrt{\frac uv},\qquad y=\sqrt{uv},

for $u,v>0$ . The region becomes the rectangle

1\le u\le4,\qquad1\le v\le4.

The forward Jacobian is

\frac{\partial(u,v)}{\partial(x,y)} =\det \begin{pmatrix} y&x\\[2pt] -y/x^2&1/x \end{pmatrix} =\frac yx+\frac yx=2v.

Hence

dA =\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv =\frac1{2v}\,du\,dv.

Since $x^2=u/v$ and $y^2=uv$ ,

4x^2-2y^2=\frac{4u}{v}-2uv.

Therefore

\begin{aligned} I &=\int_1^4\int_1^4 \left(\frac{4u}{v}-2uv\right)\frac1{2v}\,dv\,du\\ &=\int_1^4\int_1^4\left(\frac{2u}{v^2}-u\right)\,dv\,du\\ &=\int_1^4\left[-\frac{2u}{v}-uv\right]_{1}^{4}\,du\\ &=\int_1^4-\frac{3u}{2}\,du =-\frac{45}{4}. \end{aligned}

The answer is negative because the integrand takes both signs; a double integral is not automatically a geometric volume.

[challenge]

[1]:Choose and apply a coordinate change

Change of variables and polar coordinates

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