Question

How do you find the area bounded by `x=8+2y-y^2`, the y axis, y=-1, and y=3?

Answer 1

`64/3`

Explanation:

First find the integral of the function:
`int(8+2y-y^2)dy = -1/3y^3+y^2 +8y`

There are three intervals that you should solve that is `[-3,-2],[-2,0],[0,1]`.
The problem of plugging `-3` directly into the integral would subtract the area the left of the `-2` root.

(1)`int[-2,0]`
`=8(-2)+(-2)^2-1/3(-2)^3=abs(-28/3)=28/3`

(2)`int[-3,-2]=int[-3,0]-int[-2,0]`
`=8(-3)+(-3)^2-1/3(-3)^3-(-28/3)=-6-(-28/3)=10/3`

(3)`int[0,1]`
`=8(1)-(1)^2-1/3(1)^3=26/3`

Adding all together:
`28/3+10/3+26/3=64/3`