How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

1 Answer
Sep 29, 2015

See the explanation below.

Explanation:

Here is a picture of the region and a vertical slice.

enter image source here

The picture is set up to use washers (disks).
Thickness is dxdx
xx values go from -11 to 11
the radius of the larger washer is the greater yy minus -22 (the line we are revolving about is y=-2y=2)
R = 1-(-2) = 3R=1(2)=3
the radius of the smaller washer is the lesser yy minus -22
r=x^2-(-2) = x^2+2r=x2(2)=x2+2

The representative slice has volume pi(R^2-r^2)dxπ(R2r2)dx.

We need to evaluate the integral

int_-1^1 pi((3^2-(x^2+2)^2)dx=pi int_-1^1(5-4x^2-x^4)dx11π((32(x2+2)2)dx=π11(54x2x4)dx

= 104/15pi=10415π

Shells Method
If we had taken a slice horizontally:

enter image source here

This is set up to use cylindrical shells of thickness dydy

The volume of each shell is 2pi("radius")("height")("thickness")2π(radius)(height)(thickness) with thickness dydy, that is 2pi*rh*dy2πrhdy
In the region, the yy values go from 00 to 11.
The radius of the representative shell is y+2y+2 (dotted line in picture)
The height goes from the greater xx (the one on the right) to the lesser xx (the one on the left).
We need to rewrite the boundary as functions of yy instead of xx. y=x^2y=x2 becomes the two functions x=-sqrtyx=y (on the left) and x=sqrtyx=y (on the right). The height of the shell is sqrty-(-sqrty) = 2sqrtyy(y)=2y

The representative shell has volume 2pi(y+2)(2sqrty)dy2π(y+2)(2y)dy.

The solid has volume

V = int_0^1 2pi(y+2)(2sqrty)dy = 4piint_0^1 (y^(3/2)+2y^(1/2))dyV=102π(y+2)(2y)dy=4π10(y32+2y12)dy

=4pi(26/15) = 104/15 pi=4π(2615)=10415π