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The integral function is$2{鈭�}_0^{\frac{\mathrm{蟺}}{2}}鈥�{鈭�}_0^{\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}$

The goal of this issue is to sketch the region and assess the expression using polar coordinates $2{鈭�}_0^{\frac{\mathrm{蟺}}{2}}鈥�{鈭�}_0^{\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}$

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

Use the table above to draw the region ,

elaborating the expressions,

localid="1652361900299" $\begin{array}{r}2{鈭�}_0^{\mathrm{蟺}/2}鈥�{鈭�}_0^{\sin 鈦�3\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}=2{鈭�}_0^{\mathrm{蟺}/2}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sin 鈦�3\mathrm{胃}}\mathrm{诲胃}\\ =2{鈭�}_0^{\mathrm{蟺}/2}鈥�\left[\frac{{\sin }^2鈦�3\mathrm{胃}鈭�0}{2}\right]\mathrm{诲胃}\\ =2{鈭�}_0^{\mathrm{蟺}/2}鈥�\left[\frac{(1鈭�\cos 鈦�6\mathrm{胃})}{4}\right]\mathrm{诲胃}\end{array}$

We can integrating in terms of localid="1652361916107" $\mathrm{胃},$

localid="1652361908352" $2{鈭�}_0^{\mathrm{蟺}/2}鈥�{鈭�}_0^{\sin 鈦�3\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}=2{\left[\frac{\{\mathrm{胃}鈭�(\sin 鈦�6\mathrm{胃})/6\}}{4}\right]}_0^{\mathrm{蟺}/2}\left[鈭�\cos 鈦�\mathrm{xdx}=\sin 鈦�\mathrm{x}\right]$

Substituting the limits of derivatives,

localid="1652361934374" $$\begin{gathered} 2{鈭�}_0^{\mathrm{蟺}/2}鈥�{鈭�}_0^{\sin 鈦�3\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}=2\left[\frac{\{{\displaystyle \frac{\mathrm{蟺}/2鈭�(\sin 鈦�3\mathrm{蟺})}{6}}鈭�0\}}{4}\right] \\ =\frac{\mathrm{蟺}}{4} \end{gathered}$$

As a result, the integral value is localid="1652361944670" $2{鈭�}_0^{\mathrm{蟺}/2}鈥�{鈭�}_0^{\sin 鈦�3\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}=\frac{\mathrm{蟺}}{4}$