Q. 33 - Calculus

Question

Evaluate the triple integrals over the specified rectangular solid region.

$\begin{array}{r}{\iiint }_{\mathcal{R}} z\sin x\cos ydV,\text{ where }\\ \mathcal{R}=\left\{(x,y,z)\mid 0\le x\le \pi ,\frac{3\pi }{2}\le y\le 2\pi ,\text{ and }1\le z\le 3\right\}\end{array}$

Step-by-Step Solution

Answer

${\iiint }_{\mathcal{R}} z\sin x\cos ydV=8$

Step 1. Given information.

We have been given the triple integral:

$\begin{array}{r}{\iiint }_{\mathcal{R}} z\sin x\cos ydV,\text{ where }\\ \mathcal{R}=\left\{(x,y,z)\mid 0\le x\le \pi ,\frac{3\pi }{2}\le y\le 2\pi ,\text{ and }1\le z\le 3\right\}\end{array}$

We have to evaluate this over the specified rectangular solid regions.

Step 2. Evaluate.

By Fubini's theorem of triple integral :

$\begin{array}{r}{\iiint }_R z\sin x\cos ydV={\int }_0^{\pi /2\pi } {\int }_{\frac{3\pi }{2}}^3 {\int }_1^3 z\sin x\cos dxdydz\\ ={\int }_0^{\pi } {\int }_{\frac{3\pi }{2}}^{2\pi } \left[{\int }_1^3 z\sin x\cos ydz\right]dydx\\ ={\int }_0^{\pi } {\int }_{\frac{3\pi }{2}}^{2\pi } \left[\sin x\cos y{\int }_1^3 zdz\right]dydx\\ ={\int }_0^{\pi } {\int }_{\frac{3\pi }{2}}^{2\pi } \left[\sin x\cos y{\left[\frac{z^2}{2}\right]}_1^3\right]dydx\\ ={\int }_0^{\pi } {\int }_{\frac{3\pi }{2}}^{2\pi } \left[\sin x\cos y\left[\frac{3^2}{2}-\frac{1^2}{2}\right]dydx\right. \\ ={\int }_0^{\pi } {\int }_{\frac{3\pi }{2}}^{2\pi } [4\sin x\cos y]dydx\end{array}$

Step 3. Integrate with respect to y.

Integrate with respect to y

$\begin{array}{r}={\int }_0^{\pi } \left[4\sin x{\int }_{\frac{3\pi }{2}}^{2\pi } [\cos y]dy\right]dx\\ ={\int }_0^{\pi } \left[4\sin x[\sin y{]}_{\frac{3\pi }{2}}^{2\pi }\right]dx\\ ={\int }_0^{\pi } \left[4\sin x\left[\sin 2\pi -\sin \frac{3\pi }{2}\right]\right]dx\\ ={\int }_0^{\pi } [4\sin x[0-(-1)\left]\right]dx \\ ={\int }_0^{\pi } [4\sin x]dx \\ =4\left[{\int }_0^{\pi } (\sin x)\right] \\ =4\left[\right(-\cos x){]}_0^{\pi }\\ =-4\left[\right(\cos \pi )-(\cos 0\left)\right] \\ =-4[-1-1] \\ =-4[-2]\\ =8\end{array}$