# Use a double integral to find the area of the region. The region inside the circle (x-5)^2+y^2=25 and outside the circle x^2+y^2=25.

This question aims to find the area bounded by two circles using the double integral.

A bounded region is defined by a boundary or by a set of constraints. More specifically, a bounded region cannot be regarded as an infinitely large area it is usually determined by a set of parameters or measurements.

The area of a region, the volume under the surface, and the average value of the function of two variables over a rectangular region are determined by double integral. The surface integral can be referred to as a generalization of the double integral. There are two types of regions for which the area can be calculated. The first one is the Type I region which is bounded by the lines $x=a$ and $x=b$ as well as the curves $y=g(x)$ and $y=h(x)$ with the assumption that $g(x)<h(x)$ and $a<b$. In this case, the integration can be performed as: $\int\int_{R}f(x,y)dA=\int\limits_{x=a}^{b}\int\limits_{y=g(x)}^{h(x)}f(x,y)dydx$.

The second one is the Type II region which is bounded by the lines $y=c$ and $y=d$ as well as the curves $x=g(y)$ and $x=h(y)$ with the assumption that $g(y)<h(y)$ and $c<d$. In this case, the integration can be performed as: $\int\int_{R}f(x,y)dA=\int\limits_{y=c}^{d}\int\limits_{x=g(y)}^{h(y)}f(x,y)dxdy$.

To better understand the problem, the two circles are drawn and the required area is shaded in the following figure.

First, convert both the equations to the polar form. Since:

$x=r\cos\theta$  and  $y=r\sin\theta$, therefore, for $(x-5)^2+y^2=25$  we have:

$(r\cos\theta-5)^2+(r\sin\theta)^2=25$

$r^2\cos^2\theta-10r\cos\theta+25+r^2\sin^2\theta=25$

$r^2-10r\cos\theta=0$

$r^2=10r\cos\theta$

$r=10\cos\theta$                                 (1)

And for $x^2+y^2=25$, we have:

$r^2\cos^2\theta+r^2\sin^2\theta=25$

$r^2=25$

$r=5$                                                       (2)

Now, equate (1) and (2) to find the limits of integration:

$5=10\cos\theta$

$1=2\cos\theta$

$\cos\theta=\dfrac{1}{2}$

Or  $\theta=\pm\, \dfrac{\pi}{3}$

Now, set up the integral to find the area of the region as:

$\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int\limits_{5}^{10\cos\theta}rdrd\theta$

First, performing integration with respect to $r$:

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left|\dfrac{r^2}{2}\right|_{5}^{10\cos\theta}\,d\theta$

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[\dfrac{(10\cos\theta)^2}{2}-\dfrac{(5)^2}{2}\right]\,d\theta$

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[\dfrac{100\cos^2\theta}{2}-\dfrac{25}{2}\right]\,d\theta$

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[50\cos^2\theta-\dfrac{25}{2}\right]\,d\theta$

Now since $\cos^2\theta=\dfrac{\cos2\theta+1}{2}$, therefore:

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[50\left(\dfrac{\cos2\theta+1}{2}\right)-\dfrac{25}{2}\right]\,d\theta$

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[25\cos2\theta+25-\dfrac{25}{2}\right]\,d\theta$

$=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[25\cos2\theta+\dfrac{25}{2}\right]\,d\theta$

$=25\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left[\cos2\theta+\dfrac{1}{2}\right]\,d\theta$

$=25\left[\dfrac{\sin2\theta}{2}+\dfrac{\theta}{2}\right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}$

$=\dfrac{25}{2}\left[\sin\left(\dfrac{2\pi}{3}\right)+\left(\dfrac{\pi}{3}\right)-\sin\left(-\dfrac{2\pi}{3}\right)-\left(-\dfrac{\pi}{3}\right)\right]$

$=\dfrac{25}{2}\left[\dfrac{\sqrt{3}}{2}+\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}+\dfrac{\pi}{3}\right]$

$=\dfrac{25}{2}\left[\sqrt{3}+\dfrac{2\pi}{3}\right]$

$=\dfrac{25\sqrt{3}}{2}+\dfrac{25\pi}{3}$

Hence, area of the region inside the circle $(x-5)^2+y^2=25$ and outside the circle $x^2+y^2=25$ is $\dfrac{25\sqrt{3}}{2}+\dfrac{25\pi}{3}$.

## Example 1

Evaluate the double integral $\int\limits_{-1}^{1}\int\limits_{2}^{3}\dfrac{x}{y^3}\, dx dy$.

### Solution

Rewrite the integral as:

$\int\limits_{-1}^{1}\int\limits_{2}^{3}\left(\dfrac{x}{y^3}\, dx\right) dy$

Or, $\int\limits_{-1}^{1}\dfrac{1}{y^3}\left(\int\limits_{2}^{3}x\, dx\right) dy$

$=\int\limits_{-1}^{1}\dfrac{1}{y^3}\left(\left[\dfrac{x^2}{2}\right]_{2}^{3}\right) dy$

$=\int\limits_{-1}^{1}\dfrac{1}{y^3}\left[\dfrac{(3)^2}{2}-\dfrac{(2)^2}{2}\right]dy$

$=\int\limits_{-1}^{1}\dfrac{1}{y^3}\left[\dfrac{9}{2}-2\right]dy$

$=\int\limits_{-1}^{1}\dfrac{1}{y^3}\left[\dfrac{5}{2}\right]dy$

$=\dfrac{5}{2}\int\limits_{-1}^{1}\dfrac{1}{y^3}dy$

$=\dfrac{5}{2}\left[-\dfrac{1}{2y^2}\right]_{-1}^{1}$

$=\dfrac{5}{2}\left[-\dfrac{1}{2(1)^2}+\dfrac{1}{2(-1)^2}\right]$

$=\dfrac{5}{2}\left[-\dfrac{1}{2}+\dfrac{1}{2}\right]$

$=\dfrac{5}{2}(0)$

$=0$

## Example 2

Evaluate the double integral $\int\limits_{0}^{1}\int\limits_{3}^{4}x^2y\, dx dy$.

### Solution

Rewrite the integral as:

$\int\limits_{0}^{1}\int\limits_{3}^{4}\left(x^2y\, dx\right) dy$

Or, $\int\limits_{0}^{1}y\left(\int\limits_{3}^{4}x^2\, dx\right) dy$

$=\int\limits_{0}^{1}y\left(\left[\dfrac{x^3}{3}\right]_{3}^{4}\right) dy$

$=\int\limits_{0}^{1}y\left[\dfrac{(4)^3}{3}-\dfrac{(3)^3}{3}\right]dy$

$=\int\limits_{0}^{1}y\left[\dfrac{64}{3}-9\right]dy$

$=\int\limits_{0}^{1}y\left[\dfrac{37}{3}\right]dy$

$=\dfrac{37}{3}\int\limits_{0}^{1}y\,dy$

$=\dfrac{37}{3}\left[\dfrac{y^2}{2}\right]_{0}^{1}$

$=\dfrac{37}{3}\left[\dfrac{(1)^2}{2}-\dfrac{(0)^2}{2}\right]$

$=\dfrac{37}{3}\left[\dfrac{1}{2}-0\right]$

$=\dfrac{37}{3}\left[\dfrac{1}{2}\right]$

$=\dfrac{37}{6}$

Images/mathematical drawings are created with GeoGebra.