Double Integrals in Polar Coordinates

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Objectives:

1.: Be comfortable working in polar coordinates.
2.: Know how to transform a double integral in Cartesian coordinates into a double integral in polar coordinates.
3.: Understand what the scaling factor is when moving to polar coordinates.

Recap Video

The following video recaps the ideas of the section.

To recap:

Suppose $f(x,y)$ is a function, and $R$ is a region in the $xy$ -plane corresponding to a region $D$ in polar coordinates. Then

$\iint _R f(x,y)dA = \iint _D f(r\cos (\theta ),r\sin (\theta ))r dr\,d\theta .$

Example Video

Below is a video working through an example of double integrals in polar coordinates.

If $D$ is the region $x^2+y^2 \leq 4$ , $y \geq 1$ , evaluate

$\iint _D ydA.$

Compute the integral of $f(x,y) = x+y$ over the region $R$ which is the portion of $x^2+y^2 \leq 4$ in the first quadrant using polar coordinates.

Steps:

Draw the region. The region looks like:
Write the integrand $f(x,y)$ in polar coordinates: $f(r\cos (\theta ),r\sin (\theta )) = \answer {r\cos (\theta )+r\sin (\theta )}$
The region $R$ in polar coordinates can be described as: $\answer {0} \leq r \leq \answer {2} \quad \quad \quad \text { and } \quad \quad \quad 0 \leq \theta \leq \answer {\pi /2}$
The double integral in polar is therefore: $\iint _R f(x,y)dA = \int _{\answer {0}}^{\answer {\pi /2}} \int _{\answer {0}}^{\answer {2}} \answer {(r\cos (\theta )+r\sin (\theta ))r} dr\,d\theta .$
Evaluate the integral to get: $\answer {\frac {16}{3}}$

Compute the integral of $f(x,y) = x$ over the region bounded by $y = \sqrt {1-x^2}$ , $y=x$ , and $y=-x$ .

Again, draw the region, which is shown in the figure.

After drawing the region:

Transform the integrand. In polar, $f(r\cos (\theta ),r\sin (\theta )) = \answer {r\cos (\theta )}.$
Figure out the bounds for $r$ and $\theta$ . Again, drawing rays out from the origin, we see $r$ goes from the origin to the outer circle, which has polar equation $r=1$ . Therefore $\answer {0} \leq r \leq \answer {1}.$ Also note that $\theta$ goes from the line $y=x$ to the line $y=-x$ . The line $y=x$ has polar equation $\theta = \answer {\pi /4}$ (give answer between $0$ and $2\pi )$ , and the line $y=-x$ has polar equation $\theta =\answer {3\pi /4}$ .
The setup of the double integral in polar coordinates should be: $\int _{\answer {\pi /4}}^{\answer {3\pi /4}} \int _{\answer {0}}^{\answer {1}} \answer {r^2\cos (\theta )} dr\,d\theta .$
Evaluating this integral, we get: $\answer {0}$

Compute the integral of $f(x,y) = \frac {1}{\sqrt {x^2+y^2}}$ over the region inside the circle of radius $1$ centered at $(1,0)$ .

The setup of the double integral in polar coordinates is:

$\int _{\answer {-\pi /2}}^{\pi /2} \int _{\answer {0}}^{\answer {2\cos (\theta )}} \answer {1} dr\,d\theta .$ This integral evaluates to: $\answer {4}$

Compute the area of the region outside $x^2+y^2=1$ and inside $x^2+(y-1)^2=1$ .

The graph of the region is shown below.

To get the area, we want to integrate $f(x,y) = \answer {1}$ , which in polar is

$f(r\cos (\theta ),r\sin (\theta )) = \answer {1}.$ To get the bounds for $r$ and $\theta$ , first draw rays from the origin outward. Notice that they enter the region at the circle $x^2+y^2=1$ , which has polar equation $r= \answer {1}$ , and the leave the region at the circle $x^2+(y-1)^2=1$ , which has polar equation $r = \answer {2\sin (\theta )}$ . This gives us the $r$ bounds. To get $\theta$ , notice that $\theta$ will sweep between the two intersection points of the two circles. They intersect when $r = 2\sin (\theta )$ intersects $r=1$ , so when $2\sin (\theta ) = 1$ . This gives the values $\theta = \pi /6$ and $\theta = \answer {5\pi /6}$ . Therefore, the setup for the double integral is

$\int _{\pi /6}^{\answer {5\pi /6}} \int _{\answer {1}}^{\answer {2\sin (\theta )}} \answer {r}dr\,d\theta .$ This evaluates to $\answer {\frac {\sqrt {3}}{2}+\frac {\pi }{3}}$ .

Set up the following double integral in polar coordinates: $\displaystyle \int _0^1 \int _x^{\sqrt {3}x}x dy\,dx$ .
The integral becomes:

$\int _{\answer {\pi /4}}^{\answer {\pi /3}} \int _{\answer {0}}^{\answer {\sec (\theta )}}\answer {r^2 \cos (\theta )}dr\,d\theta .$

Find the volume of the solid inside $x^2+y^2=1$ bounded above by $z=x$ and below by the $xy$ -plane.

If we are going to evaluate this as a double integral $\iint _R f(x,y)dA$ , we want to let $f(x,y) = \answer {x}$ . The region in the $xy$ -plane we want is the portion of $x^2+y^2 \leq 1$ where $x \geq \answer {0}$ , which can be described in polar as:

$\answer {0} \leq r \leq \answer {1} \quad \quad \quad \text { and } \quad \quad \quad \answer {-\pi /2} \leq \theta \leq \answer {\pi /2}$ The double integral therefore becomes:

$\int _{\answer {-\pi /2}}^{\answer {\pi /2}} \int _{\answer {0}}^{\answer {1}} \answer {r^2\cos (\theta )} dr d\theta ,$ which evaluates to: $\answer {2/3}$

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