Question

In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region$\Omega =\left\{\left(x,y\right)\right|0\le x\le \frac{\mathrm{\pi }}{4}$and$\sin x\le y\le \cos x\}$$f(x,y)=x^{2}y.$

Short answer

Volume bounded by given function is:$\frac{{\mathrm{\pi }}^2-8}{64}$

Step-by-step solution

Step 1. Given Information

Function,$\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2\mathrm{y}$

Region,$\Omega =\left\{\left(x,y\right)\right|0\le x\le \frac{\mathrm{\pi }}{4}$and$\sin x\le y\le \cos x\}$

Step 2. Calculating the volume of the solid bounded above by the given function and region

The double integral is given by,

$$\begin{gathered} {\int }_0^{\frac{\pi }{4}}{\int }_{sinx}^{cosx}f(x,y)dydx \\ ={\int }_0^{\frac{\pi }{4}}{\int }_{sinx}^{cosx}{x}^{2}ydydx \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\left[{\overline{)\frac{{\mathrm{x}}^2{\mathrm{y}}^2}{2}}}_{\sin \mathrm{x}}^{\cos \mathrm{x}}\right]\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\left[\frac{{\mathrm{x}}^2({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x})}{2}\right]\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\left[\frac{{\mathrm{x}}^2(\cos 2\mathrm{x})}{2}\right]\mathrm{dx} \end{gathered}$$

Integrating by parts, we get,

$$\begin{gathered} {\overline{)\left(\frac{{\mathrm{x}}^2\sin 2\mathrm{x}}{4}+\frac{\mathrm{x}\cos 2\mathrm{x}}{4}-\frac{\sin 2\mathrm{x}}{8}\right)}}_0^{\frac{\mathrm{\pi }}{4}}=\frac{{\mathrm{\pi }}^2}{64}+0-\frac{1}{8}-0+0-0 \\ =\frac{{\mathrm{\pi }}^2-8}{64} \end{gathered}$$