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 #