`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...

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. 

...

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)`

Adding it all together

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