How do you find the volume of a solid that is enclosed by y=1/x, x=1, x=3, y=0 revolved about the y axis?

1 Answer
Jul 18, 2018

4pi

Explanation:

The shape that we are integrating should look something like a big top for a circus. We may have to split up the top and the bottom halves at y= 1/3.

We can use an integration over y here.

V = int_0^1 A(y)dy

To find the area of a slice we look at points below y=1/3 and points above.

Below
The outer radius is constant at 3. There is an inner radius cut out from the x=1 boundary, i.e. the area is the annulus between r = 1 and r = 3. We know this area:
A = pi * 3^2 - pi * 1^2 = 8pi

Above
The bound is a little more complicated above y = 1/3.
In this region, the inner radius is still a constant (hence a -pi term we will add) but the outer radius goes like 1/y, i.e.
A(y) = pi x^2 - pi = pi/y^2 - pi

We can integrate this whole thing piecewise:
V = int_0^1 A(y) dy = int_0^(1/3) A(y) dy + int_(1/3)^1 A(y)dy
int_0^(1/3) A(y) dy = 1/3 * 8pi = (8pi)/3

int_(1/3)^1 A(y) dy = -piy^-1 - pi y = -pi(1 - 3) - pi(1 - 1/3) = 2pi - 2/3 pi = (4pi)/3

Therefore, the total volume is
V = (8pi)/3 + (4pi)/3 = 4pi