# If R = (x, y): -5 less than or equal to x less than or equal to 5, -8 less than or equal to y...

## Question:

If {eq}R = \left \{ (x, y) \mid \; -5 \leq x \leq 5, \; -8 \leq y \leq 8 \right \}, {/eq} evaluate {eq}\iint_{R} \sqrt{25 - x^2} \, \mathrm{d}A {/eq}.

## Trigonometric Substitution:

Trigonometric substitution is a technique utilized to integrate functions involving {eq}a^{2} \pm x^{2} {/eq} or {eq}x^{2}-a^{2} {/eq}, where {eq}a {/eq} is a constant. A general rule of thumb is the following:

{eq}\begin{align} \displaystyle If\:\:the\:integral\:contains\hspace{10mm}\displaystyle &substitution\\ a^{2}-x^{2}\hspace{30mm} \displaystyle &x=a\sin(\theta)\\ a^{2}+x^{2}\hspace{30mm} \displaystyle &x=a\tan(\theta)\\ x^{2}-a^{2}\hspace{30mm} \displaystyle &x=a\sec(\theta)\\ \end{align} {/eq}

Step 1. Express the double integral over the described region.

{eq}\displaystyle \iint_{R} \sqrt{25 - x^2} \, \mathrm{d}A=\int_{-8}^{8}\int_{-5}^{5}\sqrt{25 - x^2}\:dx\:dy\\ {/eq}

Step 2. Evaluate the inner integral utilizing trigonometric substitution.

Let {eq}5\sin(\theta)=x {/eq}:

{eq}\displaystyle 5\cos(\theta)=dx\hspace{10mm}5\cos(\theta)=\sqrt{25-x^{2}}\hspace{10mm}\sin{-1}\left(\frac{x}{5}\right)=\theta\\ {/eq}

Calculate the boundaries of the inner integral in terms of (\theta):

When {eq}x=-5 {/eq}:

{eq}\begin{align} \displaystyle \theta\displaystyle &=\sin{-1}\left(\frac{-5}{5}\right)\\ \displaystyle &=\sin^{-1}(-1)\\ \displaystyle &=-\frac{\pi}{2}\\ \end{align} {/eq}

When {eq}x=5 {/eq}:

{eq}\begin{align} \displaystyle \theta\displaystyle &=\sin{-1}\left(\frac{5}{5}\right)\\ \displaystyle &=\sin^{-1}(1)\\ \displaystyle &=\frac{\pi}{2}\\ \end{align} {/eq}

Express the inner integral in terms of {eq}\theta {/eq}:

{eq}\begin{align} \displaystyle \int_{-8}^{8}\int_{-5}^{5}\sqrt{25 - x^2}\:dx\:dy\displaystyle &=\int_{-8}^{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 5\cos(\theta)\left(5\cos(\theta)\right)\:d\theta\:dy\\ \displaystyle &=\int_{-8}^{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}25\cos^{2}(\theta)\:d\theta\:dy\\ \end{align} {/eq}

Utilize the half angle formula and substitute {eq}\frac{1}{2}+\frac{1}{2}\cos(2\theta) {/eq} for {eq}\cos^{2}(\theta) {/eq}:

{eq}\begin{align} \displaystyle \int_{-8}^{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}25\cos^{2}(\theta)\:d\theta\:dy\displaystyle &= \int_{-8}^{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}25\left(\frac{1}{2}+\frac{1}{2}\cos(2\theta)\right)\:d\theta\:dy\\ \displaystyle &= \int_{-8}^{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{25}{2}+\frac{25}{2}\cos(2\theta)\right)\:d\theta\:dy\\ \displaystyle & =\int_{-8}^{8}\left[\frac{25}{2}\theta-\frac{25}{4}\sin(2\theta)\right]_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}dy\\ \displaystyle &=\int_{-8}^{8}\left[\left(\frac{25\pi}{4}-0\right)-\left(-\frac{25\pi}{4}-0\right)\right]\:dy\\ \displaystyle &=\int_{-8}^{8}\left(\frac{25\pi}{4}+\frac{25\pi}{4}\right)\:dy\\ \displaystyle &=\int_{-8}^{8}\frac{25\pi}{2}\:dy\\ \end{align} {/eq}

Step 3. Evaluate the outer integral.

{eq}\begin{align} \displaystyle \int_{-8}^{8}\frac{25\pi}{2}\:dy\displaystyle &=\left[\frac{25\pi}{2}y\right]_{y=-8}^{y=8}\\ \displaystyle &=\frac{25\pi}{2}(8)-\left(\frac{25\pi}{2}(-8)\right)\\ \displaystyle &=100\pi+100\pi\\ \displaystyle &=200\pi\\ \end{align} {/eq}