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

May 16, 2017

$\frac{64}{3}$

#### Explanation:

First find the integral of the function:
$\int \left(8 + 2 y - {y}^{2}\right) \mathrm{dy} = - \frac{1}{3} {y}^{3} + {y}^{2} + 8 y$

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

(1)$\int \left[- 2 , 0\right]$
$= 8 \left(- 2\right) + {\left(- 2\right)}^{2} - \frac{1}{3} {\left(- 2\right)}^{3} = \left\mid - \frac{28}{3} \right\mid = \frac{28}{3}$

(2)$\int \left[- 3 , - 2\right] = \int \left[- 3 , 0\right] - \int \left[- 2 , 0\right]$
$= 8 \left(- 3\right) + {\left(- 3\right)}^{2} - \frac{1}{3} {\left(- 3\right)}^{3} - \left(- \frac{28}{3}\right) = - 6 - \left(- \frac{28}{3}\right) = \frac{10}{3}$

(3)$\int \left[0 , 1\right]$
$= 8 \left(1\right) - {\left(1\right)}^{2} - \frac{1}{3} {\left(1\right)}^{3} = \frac{26}{3}$

$\frac{28}{3} + \frac{10}{3} + \frac{26}{3} = \frac{64}{3}$