Question

Find the volume of the solid bounded above by the given function over the specified region

$f(x,y)=10-2x+y$, with $\Omega$ the region from Exercise 21

Short answer

The volume is :

$10e+\frac{e^2}{4}-\frac{49}{4}$cubic units.

Step-by-step solution

Step 1. Given information

The function:

$f(x,y)=10-2x+y$

The region in execise 21

Step 2. To setup integral to find volume

In the given region we can see that

$$\begin{gathered} 0\le y\le e^x \\ 0\le x\le 1 \end{gathered}$$

So volume of the function over this region is:

$$\begin{gathered} V={\int }_0^1{\int }_0^{e^x}f(x,y)dydx \\ ={\int }_0^1{\int }_0^{e^x}\left(10-2x+y\right)dydx \end{gathered}$$

Step 3. To evaluate integral.

First integrate the function with respect to y.

$$\begin{gathered} {\int }_0^{e^x}\left(10-2x+y\right)dy \\ ={\left[10y-2xy+\frac{y^2}{2}\right]}_0^{e^x} \\ =\left(10e^x-2x{e}^x+\frac{e^{2x}}{2}\right)-0 \\ =10e^x-2x{e}^x+\frac{e^{2x}}{2} \end{gathered}$$

Now integrate with respect to x.

$$\begin{gathered} {\int }_0^1\left(10e^x-2x{e}^x+\frac{e^{2x}}{2}\right)dx \\ ={\left[10e^x-2\left(x{e}^x-e^x\right)+\frac{e^{2x}}{4}\right]}_0^1 \\ =\left(10e-2(e-e)+\frac{e^2}{4}\right)-\left(10-2(0-1)+\frac{1}{4}\right) \\ =10e-0+\frac{e^2}{4}-10-2-\frac{1}{4} \\ =10e+\frac{e^2}{4}-\frac{49}{4} \end{gathered}$$