Q. 39 - Calculus

Question

In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region

$f(x, y)=\sin x \cos y$and $\mathrm{\Omega }=\left\{\right(x,y\left)\right|-2\le x\le 2$ and $\left|x\right|\le y\le 2\}$

Step-by-Step Solution

Answer

The volume of the described solid is: 0

Step 1. Given Information

A function, $f(x, y)=\sin x \cos y$

The region, $\mathrm{\Omega }=\left\{\right(x,y\left)\right|-2\le x\le 2$ and $\left|x\right|\le y\le 2\}$

Step 2. Finding the volume of given solid

The double integral to find the volume of the given solid is given by,

$$\begin{gathered} {\int }_{\left|x\right|}^2{\int }_{-2}^2\left(\sin x \right)\left(\cos y\right) dx dy \\ ={\int }_{\left|x\right|}^2{\int }_{-2}^2\left(\sin x \right)\left(\cos y\right) dx dy \\ ={\int }_{\left|x\right|}^2\left(\cos y\right)\left[{\overline{)\left(-\cos x \right)}}_{-2}^2\right] dy \\ ={\int }_{\left|x\right|}^2\left(\cos y\right)\left[-\cos \left(2\right)+cos (-2) \right] dy \\ ={\int }_{\left|x\right|}^2\left(\cos y\right)\left[-\cos \left(2\right)+cos \left(2\right)\right] dy \\ ={\int }_{\left|x\right|}^2\left(\cos y\right)\left[0\right] dy \\ =0 \end{gathered}$$