Q.28 - Calculus

Question

Each of the integrals or integral expressions in Exercises 28 represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.

$2{\int }_0^{2\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta -2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta$

Step-by-Step Solution

Answer

The value of the integral is

$2{\int }_0^{2\pi /3}{\int }_0^{(1/2)rces\theta }rdrd\theta -2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)\cos \theta }rdrd\theta =\frac{\pi }{12}$

Step 1: Given information

The expression is

$2{\int }_0^{2\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta -2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta$

Step 2: Simplification

Here, $r=0,r=(1/2)+\cos \theta$ and $\theta =0,\theta =2\pi /3,\theta =\pi$ and $\theta =4\pi /3$

$\begin{array}{|ll|}\hline \theta & r=(1/2)+\cos \theta \\ 0& 1.5\\ \pi /4& 1.2071\\ \pi /2& 0.5\\ 3\pi /4& -0.2071\\ \pi & -0.5\\ 5\pi /4& -0.2071\\ 4\pi /3& 0\\ \hline\end{array}$

To sketch the region use the above table

 Plot of $r=(1/2)+\cos \theta$

$I=2{\int }_0^{2\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta -2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)+sen\theta }rdrd\theta$

$I=I_1-I_2$

Where,

$$\begin{gathered} I_1=2{\int }_0^{2\pi /3}{\int }_0^{(1/2)+eos\theta }rdrd\theta \\ {I}_1=2{\int }_0^{2\pi /3}{\left[\frac{r^2}{2}\right]}_0^{(1/2)+\cos \theta }d\theta \\ {I}_1=2{\int }_0^{2\pi /3}\left[\frac{\left\{\right(1/2)+\cos \theta {\}}^2-0}{2}d\theta \right. \\ {I}_1=2{\int }_0^{2\pi /3}\left[\frac{\left\{\frac{1}{4}+\cos \theta +{\cos }^2\theta \right\}}{2}\right]\theta \\ {I}_1=2{\int }_0^{2\pi /3}\left[\frac{\left\{\frac{1}{4}+\cos \theta +\frac{1}{2}(1+\cos 2\theta )\right\}}{2}d\theta \right. \end{gathered}$$

Integrate with respect to $\theta$.

$$\begin{gathered} I_1=2{\left[\frac{\left\{\frac{\theta }{4}+\sin \theta +\frac{1}{2}\left(\theta +\frac{1}{2}\sin 2\theta \right)\right\}}{2}\right]}_0^{2\pi /3} \\ {I}_1=2\left[\frac{\left.\left\{\frac{3}{4}·\frac{2\pi }{3}+\sin \left(\frac{2\pi }{3}\right)+\frac{1}{4}\sin \left(\frac{4\pi }{3}\right)\right\}-\left\{0\right\}\right]}{2}\right] \\ {I}_1=2\left[\frac{\left\{\frac{\pi }{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8}\right\}}{2}\right] \\ {I}_1=2\left[\frac{\left\{\frac{\pi }{2}+\frac{3\sqrt{3}}{8}\right\}}{2}\right] \\ {I}_1=\frac{\pi }{2}+\frac{3\sqrt{3}}{8} \end{gathered}$$

Now

$$\begin{gathered} I_2=2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)+\cos \theta }rdrd\theta \\ {I}_2=2{\int }_{\pi }^{4\pi /3}{\left[\frac{r^2}{2}\right]}_0^{(1/2)+\cos \theta }d\theta \\ {I}_2=2{\int }_n^{4\pi /3}\left[\left\{\frac{\left.\frac{1}{4}+\cos \theta +{\cos }^2\theta \right\}}{2}\right]d\theta \right. \\ {I}_2=2{\int }_n^{4\pi /3}\left[\frac{\left\{\frac{1}{4}+\cos \theta +\frac{1}{2}(1+\cos 2\theta )\right\}}{2}d\theta \right. \\ {I}_2=2{\left[\frac{\left\{\frac{\theta }{4}+\sin \theta +\frac{1}{2}\left(\theta +\frac{1}{2}\sin 2\theta \right)\right\}}{2}\right]}_n^{4n} \\ {I}_2=2\left[\frac{\left\{\frac{1}{4}·\frac{4\pi }{3}+\sin \left(\frac{4\pi }{3}\right)+\frac{1}{4}\sin \left(\frac{8\pi }{3}\right)\right\}-\left\{\frac{\pi }{4}+\sin \pi +\frac{1}{2}\left(\pi +\frac{1}{2}\sin 2\pi \right)\right\}}{2}\right] \\ {I}_2=2\left[\frac{\left\{\frac{\pi }{3}-\frac{\sqrt{3}}{2}+\frac{1}{4},\frac{\sqrt{3}}{2}\right\}-\left\{\frac{\pi }{4}+\frac{\pi }{2}\right\}}{2}\right] \\ {I}_2=2\left[\frac{\left\{\frac{\pi }{3}-\frac{\sqrt{3}}{2}+\frac{1}{4}·\frac{\sqrt{3}}{2}\right\}-\left\{\frac{\pi }{4}+\frac{\pi }{2}\right\}}{2}\right] \\ {I}_2=-\frac{5\pi }{12}-\frac{3\sqrt{3}}{8} \end{gathered}$$

Therefore,

$$\begin{gathered} I=\frac{\pi }{2}+\frac{3\sqrt{3}}{8}-\frac{5\pi }{12}-\frac{3\sqrt{3}}{8} \\ I=\frac{\pi }{12} \end{gathered}$$

Thus, the value of integral is

$2{\int }_0^{2\pi /3}{\int }_0^{(1/2)+ces\theta }rdrd\theta -2{\int }_{\pi }^{4\pi /3}{\int }_0^{(1/2)res\theta }rdrd\theta =\frac{\pi }{12}$