91影视

The Given integral is$2\underset{\frac{鈭�\mathrm{蟺}}{2}}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}\underset{0}{\overset{2鈭�\mathrm{蝉颈苍胃}}{鈭�}}\mathrm{r}\text{鈥�}\mathrm{dr}\text{鈥�}\mathrm{诲胃}$

The goal of this challenge is to sketch the region in polar coordinates and assess the expression,$2\underset{\frac{鈭�蟺}{2}}{\overset{\frac{蟺}{2}}{鈭�}}\underset{0}{\overset{2鈭�\sin 胃}{鈭�}}r\text{鈥�}dr\text{鈥�}d胃$

using the values , $\mathrm{r}=0,\mathrm{r}=2鈭�\sin 鈦�\mathrm{胃}\text{and}\mathrm{胃}=鈭�\frac{{\displaystyle \mathrm{蟺}}}{{\displaystyle 2}},\mathrm{胃}=\frac{{\displaystyle \mathrm{蟺}}}{{\displaystyle 2}}$

Use the table above to draw the region

$\begin{array}{r}2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�{鈭�}_0^{2鈭�\sin 鈦�\mathrm{胃}}鈥�\mathrm{rdr诲胃}=2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{2鈭�\sin 鈦�\mathrm{胃}}\mathrm{诲胃}\\ =2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�\left[\frac{(2鈭�\sin 鈦�\mathrm{胃}{)}^2鈭�0}{2}\mathrm{诲胃}\right.\\ =2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�\left[\frac{\left(4鈭�4\sin 鈦�\mathrm{胃}+{\sin }^2鈦�\mathrm{胃}\right)}{2}\right]\mathrm{胃}\left[(\mathrm{a}鈭�\mathrm{b}{)}^2={\mathrm{a}}^2鈭�2\mathrm{ab}+{\mathrm{b}}^2\right]\\ =2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�\left[\frac{\{4鈭�4\sin 鈦�\mathrm{胃}+(1鈭�\cos 鈦�2\mathrm{胃})/2\}}{2}\right]\mathrm{胃}\\ =2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�\left[\frac{\{9鈭�8\sin 鈦�\mathrm{胃}鈭�\cos 鈦�2\mathrm{胃}\}}{4}\mathrm{诲胃}\right.\end{array}$

We can integrate in terms of this,$\mathrm{胃}$

$\begin{array}{r}2{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}鈥�{鈭�}_0^{2鈭�\sin 鈦�\mathrm{胃}}鈥�\mathrm{rdr诲胃}=2{\left[\frac{(9\mathrm{胃}+8\cos 鈦�\mathrm{胃}鈭�(\sin 鈦�2\mathrm{胃})/2\}}{4}\right]}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/2}\\ \left[鈭�\sin 鈦�\mathrm{xdx}=鈭�\cos 鈦�\mathrm{x},鈭�\cos 鈦�\mathrm{xdx}=\sin 鈦�\mathrm{x}\right]\end{array}$

$\left[鈭�\sin 鈦�\mathrm{xdx}=鈭�\cos 鈦�\mathrm{x},鈭�\cos 鈦�\mathrm{xdx}=\sin 鈦�\mathrm{x}\right]$

Substituting the limits of derivatives,

$$\begin{gathered} 2\underset{\frac{鈭�\mathrm{蟺}}{2}}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}\underset{0}{\overset{2鈭�\mathrm{蝉颈苍胃}}{鈭�}}\mathrm{r}\text{鈥�}\mathrm{dr}\text{鈥�}d胃 \\ =2\left[\frac{{\displaystyle \frac{\frac{9\mathrm{蟺}}{2}}{2}+8\cos 鈦�\left(\frac{\mathrm{蟺}}{2}\right)鈭�\frac{\left(\mathrm{蝉颈苍蟺}\right)}{2}\}鈭�\{-\frac{\frac{9\mathrm{蟺}}{2}}{2}+8\cos 鈦�\left(\frac{\mathrm{蟺}}{2}\right)+\frac{\left(\mathrm{蝉颈苍蟺}\right)}{2}\}}}{4}\right] \\ =2\left[\frac{\left\{\frac{9\mathrm{蟺}}{2}\right\}{\displaystyle -}{\displaystyle \left\{-\frac{{\displaystyle 9\mathrm{蟺}}}{{\displaystyle 2}}\right\}}}{4}\right] \\ =\frac{{\displaystyle 9\mathrm{蟺}}}{{\displaystyle 2}} \end{gathered}$$

As a result, the integral value is $2\underset{\frac{鈭�\mathrm{蟺}}{2}}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}\underset{0}{\overset{2鈭�\mathrm{蝉颈苍胃}}{鈭�}}\mathrm{r}\text{鈥�}\mathrm{dr}\text{鈥�}\mathrm{诲胃}=\frac{9\mathrm{蟺}}{2}$