Q.46 - Calculus

Question

Each of the integrals or integral expressions in Exercises 39–46 represents the volume of a solid in ${\mathrm{ℝ}}^3$. Use polar coordinates to describe the solid, 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 given expressions 

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

Step 2: Simplificaiton

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

Suppose

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

$I=I_1-I_2$

Where,

$$\begin{gathered} I_1=2{\int }_{-\pi /4}^{\pi /2}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta \\ {I}_1=2{\int }_{-\pi /4}^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{(\sqrt{2}/2)+\sin \theta }d\theta \\ {I}_1=2{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/2}\left[\frac{\left\{\right(\sqrt{2}/2)+\sin \theta {\}}^2-0}{2}\right]\theta \end{gathered}$$

Now

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

$$\begin{gathered} I_2=2{{\left[\frac{\left\{\theta -\sqrt{2}\cos \theta -\frac{1}{4}\sin 2\theta \right\}}{2}\right]}_{-\pi /2}}_{}^{-\pi /4} \\ =2\left[\frac{\left\{-\frac{\pi }{4}-\sqrt{2}\cos \left(-\frac{\pi }{4}\right)-\frac{1}{4}\sin \left(-\frac{\pi }{2}\right)\right\}-\left\{-\frac{\pi }{2}-\sqrt{2}\cos \left(-\frac{\pi }{2}\right)-\frac{1}{4}\sin (-\pi )\right\}}{2}\right] \end{gathered}$$

$I_2=2\left[\frac{\left\{-\frac{\pi }{4}-1+\frac{1}{4}\right\}-\left\{-\frac{\pi }{2}\right\}}{2}\right]$

$I_2=\frac{3}{4}(\pi -1)$

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

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