Question

(A) Find the area of the region enclosed by x=y² and y= x-2

(B) Find the volume of the solid obtained by revolving the region (part A) about the y-axis.

Answer 1

First, we can rewrite socen equations:

$y=x-2 \to x = y +2$

(A) find the intersection points of the graphs of functions:

We need to solve system:

$\{x = y^2, x=y+2\\ y^2 -y-2 = 0\\ y=2, y=-1$

And now we can find the area:

$S = \int_{-1}^2( 2+y-y^2) dy = 2y+\cfrac{y^2}{2}-\cfrac{y^3}{3}|_{-1}^2=4+2-\cfrac{8}{3}-\\ -(-2 +\cfrac{1}{2}+\cfrac{1}{3}) = 5 - \cfrac{1}{2} = 4.5$

(B)

$V = \pi\int_a^bf^2(y)dy = \pi\int_{-1}^2(2+y-y^2)^2dy = \\ =\pi\int_{-1}^2(4+2y-2y^2+y^2+2y-y^3+y^4+2y^2-y^3)dy = \\ =\pi\int_{-1}^2(4+y^2+y^4+4y-4y^2-2y^3)=\\ =\pi\int_{-1}^2(4-3y^2+y^4-2y^3)dy = \\ =\pi(4y-y^3+\cfrac{y^5}{5}-\cfrac{y^4}{2}|_{-1}^2) =\\ =\pi(8 - 8+\cfrac{32}{5}-8-(-4+1-\cfrac{1}{5}-\cfrac{1}{2})) = \\ = \pi(\cfrac{33}{5}-5+\cfrac{1}{2}) = \pi\cfrac{66-50+5}{10} = \cfrac{21}{10}\pi$