Q.27 - Calculus

Question

Each of the integrals or integral expressions in Exercises 27 represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.

$2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta -2{\int }_{-\pi /2}^{-\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta$

Step-by-Step Solution

Answer

 The value of integral is

$2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)·\sin \theta }rdrd\theta -2{\int }_{-\pi /2}^{-\pi /4}{\int }_0^{(\sqrt{2}/2)\sin \theta }rdrd\theta =\frac{3}{2}$

step 1: Given information

The expression is

$2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta -2{\int }_{-\pi /2}^{-\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta$ 

Step 2: Simplification

Here, $r=0,r=(\sqrt{2}/2)+\sin \theta$ and $\theta =-\pi /4,\theta =-\pi /2,\theta =\pi /2$

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

To sketch the region use the above table

Plot of $r=(\sqrt{2}/2)+\sin \theta$

$I=2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta -2{\int }_{-\pi /2}^{-\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta$

$I=I_1-I_2$

Where,

$I_1=2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)+sis\theta }rdrd\theta$

$I_1=2{\int }_{-\pi /4}^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{(\sqrt{2}/2)·\sin \theta }d\theta$

$I_1=2{\int }_{-\pi /4}^{\pi /2}\left[\frac{\left\{\right(\sqrt{2}/2)+\sin \theta {\}}^2-0}{2}\right]\theta$

$I_1=2{\int }_{-\pi /4}^{\pi /2}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \theta +{\sin }^2\theta \right\}}{2}d\theta \right.$

$I_1=2{\int }_{-\pi /4}^{\pi /2}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \theta +\frac{1}{2}(1-\cos 2\theta )\right\}}{2}d\theta \right.$

Integrate with respect to $\theta \text{. }$

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

now,

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

Therefore,

$$\begin{gathered} I=\frac{3}{4}(\pi +1)-\frac{3}{4}(\pi -1) \\ I=\frac{3}{2} \end{gathered}$$

Thus, the value of integral is

$2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)·\sin \theta }rdrd\theta -2{\int }_{-\pi /2}^{-\pi /4}{\int }_0^{(\sqrt{2}/2)·\sin \theta }rdrd\theta =\frac{3}{2}$