# `xy = 1, y = 0, x = 1, x = 2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical...

`xy = 1, y = 0, x = 1, x = 2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. (about x = -1)

By using Washer method, we can find the volume of the solid.

`V = pi*int_a^b(f^2 (y) - g^2 (y) dy f(y)gtg(y) `

Since the curve is bounded by x=1 and x=2, then y-values are bounded between 0 and 1. If you graph this curve and bounds, you'll see that we have to split the integral into 2 separate integrals.

First Integral

The first one will be bounded between y=0 and y = 1/2.

From y=0 to y=1/2, the area being revolved around x=-1 is just a rectangle. The outer radius is x=2 and inner radius is x=1.

So,

`V_1 = pi*int_0^(1/2) (2-(-1))^2 - (1-(-1))^2 dy `

`   = pi*int_0^(1/2) 9 - 4 dy `

`   = 5pi*1/2 `

` V_1  = 5pi/2 `

Second Integral

Now, the second region is bounded from y=1/2 to y=1 and the outer function is x=1/y and the inner function is x=1.

So,

`V_2 = pi*int_(1/2)^1 (1/y-(-1))^2 - (1-(-1))^2 dy`

`       = pi*int_(1/2)^1 1/y^2+2/y+1 - 4 dy`

`       = pi*int_(1/2)^1 1/y^2+2/y-3 dy`

`       = pi|_(1/2)^1 -1/y+2lny-3y `

`V_2 = pi(ln4 -1/2)`

Now that we have the volumes of the 2 different regions, we can add them together to get the total volume.

So, `V= V_1 + V_2 = 5pi/2 + piln4 - pi/2 = pi(2+ln4)`

`V = pi(2+ln4)` is the final answer