Q. 45 - 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.

${\int }_{-\pi /4}^{5\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta$

Step-by-Step Solution

Answer

The value of integral is 

${\int }_{-\pi /4}^{5\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta =\frac{3}{4}(\pi +1)$

Step 1: Given information

The expression is 

$I={\int }_{-\pi /4}^{5\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta$

Step 2: Calculation

Here, $r=0, r=(\sqrt{2}/2)+\sin \theta and \theta =-\pi /4 ,\theta =5\pi /4$

Integrate with respect to $r$ first,

$I={\int }_{-\pi /4}^{5\pi /4}{\left[\frac{r^2}{2}\right]}_0^{(\sqrt{2}/2)+\sin \theta }d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

Put the limits

$$\begin{gathered} I={\int }_{-\pi /4}^{5\pi /4}\left[\frac{\left\{\right(\sqrt{2}/2)+\sin \theta {\}}^2-0}{2}\right]d\theta \\ I={\int }_{-\pi /4}^{5\pi /4}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \theta +{\sin }^2\theta \right\}}{2}\right]d\theta \\ I={\int }_{-\pi /4}^{5\pi /4}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \theta +\frac{1}{2}(1-\cos 2\theta )\right\}}{2}\right]d\theta \end{gathered}$$

Now integrate with respect to$\theta$

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

$$\begin{gathered} I=\left[\frac{{\left.\left\{\theta -\sqrt{2}\cos \theta -\frac{1}{4}\sin 2\theta \right\}\right]}_{-\pi /4}^{5\pi /4}}{2}\right] \\ I=\left[\frac{\left\{\frac{5\pi }{4}-\sqrt{2}\cos \left(\frac{5\pi }{4}\right)-\frac{1}{4}\sin \left(\frac{5\pi }{2}\right)\right\}-\left\{\frac{-\pi }{4}-\sqrt{2}\cos \left(\frac{-\pi }{4}\right)-\frac{1}{4}\sin \left(\frac{-\pi }{2}\right)\right\}}{2}\right] \end{gathered}$$

$$\begin{gathered} I=\left[\frac{\left\{\frac{5\pi }{4}-\sqrt{2}\left(-\frac{1}{\sqrt{2}}\right)-\frac{1}{4}\right\}-\left\{\frac{-\pi }{4}-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)+\frac{1}{4}\right\}}{2}\right] \\ I=\left[\frac{\left\{\frac{5\pi }{4}+1-\frac{1}{4}\right\}-\left\{\frac{-\pi }{4}-1+\frac{1}{4}\right\}}{2}\right] \\ I=\frac{3}{4}(\pi +1) \end{gathered}$$

Thus, the value of integral is

${\int }_{-\pi /4}^{5\pi /4}{\int }_0^{(\sqrt{2}/2)+\sin \theta }rdrd\theta =\frac{3}{4}(\pi +1)$