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 22.

Short answer

The volume of the function is:$54$cubic inches.

Step-by-step solution

Step 1. Given information

The function:

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

The region in exercise 22

Step 2. To setup integral 

In the given region the $-x\le y\le x$and $0\le x\le 3$

So, the volume of the function over specified region is:

$$\begin{gathered} V={\int }_0^3{\int }_{-x}^{x}f(x,y)dydx \\ ={\int }_0^3{\int }_{-x}^x\left(10-2x+y\right)dydx \end{gathered}$$

Step 3. Evaluate integral

First integrate the function with respect to y.

$$\begin{gathered} {\int }_{-x}^x\left(10-2x+y\right)dy \\ ={\left[10y-2xy+\frac{y^2}{2}\right]}_{-x}^x \\ =\left(10x-2x^2+\frac{x^2}{2}\right)-\left(-10x+2x^2+\frac{x^2}{2}\right) \\ =20x-4x^2 \end{gathered}$$

Now integrate the above function of x with respect to x.

$$\begin{gathered} {\int }_0^3(20x-4x^2)dx \\ ={\left[20\frac{x^2}{2}-4\frac{x^3}{3}\right]}_0^3 \\ ={\left[10x^2-\frac{4}{3}{x}^3\right]}_0^3 \\ =\left[10{\left(3\right)}^2-\frac{4}{3}{\left(3\right)}^3-0\right] \\ =90-36 \\ =54 \end{gathered}$$