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

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. \(\displaystyle\int_0^{2\pi} \displaystyle\int_1^2 r^2\ dr\ d\theta = \dfrac{14\pi}{3}\)