$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Double Integrals with Polar Coordinates

## Introduction

There are cases where it is more convenient to use polar coordinates to evaluate a double integral. Recall that polar and rectangular coordinates are related by the following identities: \begin{align} x=r \cos \theta \ \ \ \ &\textrm{ and }\ \ \ \ y=r \sin \theta\\ r^2=x^2+y^2 \ \ \ \ &\textrm{ and }\ \ \ \ \theta = \tan^{-1} \dfrac{y}{x} \end{align} Therefore, the function $$f(x,y)$$ can be written $$f(r \cos \theta, r \sin \theta)$$: $\iint_R f(x,y)\ dA = \iint_R f(r \cos \theta, r \sin \theta)\ ?$ The question is, what is $$dA$$?

To answer this, we'll draw a small polar "rectangle" as shown below and shrink it down until the difference between the radii is a differential $$dr$$, and the difference between the angles is $$d\theta$$.

To find the area of this slice, note that the area of the large wedge is the area of the large circle times the proportion of the circle that it covers:

$A_2 = \pi r_2^2 \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right)$

The area of the small wedge is similar:

$A_1 = \pi r_1^2 \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right)$

Therefore, the area of the shaded rectangle is

$A_{slice} = \pi (r_2^2 - r_1^2) \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right) = (r_2+r_1)(r_2-r_1) \left(\dfrac{\theta_2 - \theta_1}{2}\right)$

If $$\Delta \theta = \theta_2 - \theta_1$$ and $$\Delta r = r_2 - r_1$$, then $A_{slice} = \dfrac{r_2+r_1}{2}\ \Delta r\ \Delta \theta.$ As the slice gets smaller, going toward differentials, $$r_2 \to r_1$$, so the differential area is $dA = r\ dr\ d\theta.$

Therefore, the rectangular double integral can be rewritten in polar coordinates as follows: (don't forget the $$r$$ before $$dr\ d\theta$$)

$\iint_R f(x,y)\ dA = \iint_R f(r \cos \theta, r \sin \theta)\ r\ dr\ d\theta$

What you'll see as we do examples is that we tend to use polar coordinates when we see $$x^2+y^2$$ somewhere in the double integrals, since we can replace that with $$r^2$$.

## Examples

Find the volume of the solid bounded by the paraboloid $$z=9-x^2-y^2$$ and the xy plane.

#### Solution

We notice that switching to polar coordinates will simplify the integral:

$\iint_R 9-x^2-y^2\ dA = \iint_R (9-r^2)\ r\ dr\ d\theta$

To find the limits of integration (ranges of $$r$$ and $$\theta$$), note that when the paraboloid intersects the xy plane (where $$z=0$$), $$x^2+y^2=9$$, which describes a circle centered at the origin with radius 3. Therefore, $$r$$ ranges from 0 (the center of this circle) to 3 (the edge of the circle).

$0 \leq r \leq 3\ \ \ \ \textrm{ and }\ \ \ \ 0 \leq \theta \leq 2\pi$

The double integral, then, is

\begin{align} \int_0^{2\pi} \int_0^3 9r-r^3\ dr\ d\theta &= \int_0^{2\pi} \bigg[\dfrac{9}{2}r^2 - \dfrac{1}{4}r^4 \bigg]_0^3\ d\theta\\ &= \int_0^{2\pi} \dfrac{81}{4}\ d\theta = \dfrac{81}{4}\theta \bigg|_0^{2\pi} = \ans{\dfrac{81\pi}{2}} \end{align}

Find the volume of the solid under the paraboloid $$z=x^2+y^2$$, above the xy plane, and inside the cylinder $$x^2+y^2=2x$$.

#### Solution

The function $$x^2+y^2$$ will be replaced with $$r^2$$, so all that remains is to find the limits represented by the cylinder $$x^2+y^2=2x$$:

\begin{align} x^2+y^2 &= 2x\\ r^2 &= 2 r \cos \theta\\ r(r-2\cos\theta) &= 0\\ r &= 0, 2 \cos \theta \end{align}

What about limits on $$\theta$$? Note that $$x^2+y^2=2x$$ can be rewritten $$(x-1)^2+y^2=1$$, which represents a circle of radius 1 centered at $$(1,0)$$, so

$-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$

Therefore, the double integral is

\begin{align} \iint_R x^2+y^2\ dA &= \iint_R r^2 \cdot r\ dr\ d\theta\\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r^3\ dr\ d\theta = \int_{-\pi/2}^{\pi/2} 4\cos^4 \theta\ d\theta = \ldots = \ans{\dfrac{3\pi}{2}} \end{align}

Find the volume of the solid bounded by the paraboloids $$z=x^2+y^2$$ and $$z=8-x^2-y^2$$.

#### Solution

To find the limits, find where these surfaces intersect: \begin{align} x^2+y^2 &= 8-x^2-y^2\\ 2(x^2+y^2) &= 8\\ x^2+y^2 &= 4 \end{align} which is a circle of radius 2 centered at the origin, so

$0 \leq r \leq 2 \ \ \ \ \textrm{ and }\ \ \ \ 0 \leq \theta \leq 2\pi$

Therefore, the double integral to find this volume is

\begin{align} \iint_R &(8-x^2-y^2)-(x^2+y^2)\ dA\\ &= \iint [(8-r^2)-r^2]\ r\ dr\ d\theta\\ &= \int_0^{2\pi} \int_0^2 8r-2r^3\ dr\ d\theta\\ &= \int_0^{2\pi} \bigg[ 4r^2-\dfrac{1}{2}r^4 \bigg]_0^2\ d\theta\\ &= \int_0^{2\pi} 8\ d\theta = \ans{16\pi} \end{align}

Find the volume of the solid bounded by the paraboloid $$z=4-x^2-y^2$$ above the region that is outside the circle $$r=2$$ and inside the circle $$r=4 \cos \theta$$ (these two circles are shown below).

#### Solution

The limits on $$r$$ are pretty straightforward: $2 \leq r \leq 4 \cos \theta$ but what about the limits on $$\theta$$? In other words, what is $$\theta$$ when the two circles intersect?

$4 \cos \theta = 2 \longrightarrow \cos \theta = \dfrac{1}{2} \longrightarrow \theta = \pm \dfrac{\pi}{3}$

Therefore, the volume of this solid is

\begin{align} \int_{-\pi/3}^{\pi/3} &\int_2^{4\cos \theta} (4-r^2)\ r\ dr\ d\theta\\ &= \int_{-\pi/3}^{\pi/3} \bigg[ 2r^2-\dfrac{1}{4}r^4 \bigg]_2^{4\cos\theta}\ d\theta\\ &= \int_{-\pi/3}^{\pi/3} 32 \cos^2 \theta - 64\cos^4 \theta - 4\ d\theta = \ans{-6\sqrt{3}-8\pi} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)

1. Evaluate $$\displaystyle\iint_R x^2+y^2\ dA$$, where $$R=\{(r,\theta)\ :\ 0 \leq r \leq 4, 0 \leq \theta \leq 2\pi\}$$.
2. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^4 r^3\ dr\ d\theta = 128\pi$$

3. Evaluate $$\displaystyle\iint_R 2xy\ dA$$, where $$R=\{(x,y)\ :\ x^2+y^2 \leq 9, y \geq 0\}$$.
4. $$\displaystyle\int_0^{\pi} \displaystyle\int_0^3 2r^3 \cos\theta\ \sin\theta\ dr\ d\theta = 0$$

5. Evaluate $$\displaystyle\iint_R \dfrac{1}{\sqrt{16-x^2-y^2}}\ dA$$, where $$R=\{(x,y)\ :\ x^2+y^2 \leq 4,x \geq 0, y \geq 0\}$$.
6. $$\displaystyle\int_0^{\pi/2} \displaystyle\int_0^2 \dfrac{r}{\sqrt{16-r^2}}\ dr\ d\theta = (2-\sqrt{3})\pi$$

7. Evaluate the integral $$\displaystyle\int_{-1}^1 \displaystyle\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2}\ dy\ dx$$.
8. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^1 r^4\ dr\ d\theta = \dfrac{2\pi}{5}$$

9. Evaluate the integral $$\displaystyle\int_{0}^{\pi/4} \displaystyle\int_{0}^{\sin\theta} r\ dr\ d\theta$$.
10. $$=\dfrac{1}{4}\theta - \dfrac{1}{8}\cos(2\theta) \bigg|_0^{\pi/4} = \dfrac{\pi}{16} + \dfrac{1}{8}$$

11. Evaluate $$\displaystyle\iint_R \sqrt{x^2+y^2}\ dA$$, where $$R=\{(x,y)\ :\ 1 \leq x^2+y^2 \leq 4\}$$.
12. $$\displaystyle\int_0^{2\pi} \displaystyle\int_1^2 r^2\ dr\ d\theta = \dfrac{14\pi}{3}$$