# How do you find the volume of the solid obtained by rotating the region bounded by: #y=sqrt(x-1)#, y=0, x=5 rotated about y=7?

##### 2 Answers

Please see below.

#### Explanation:

I would use shells. Here is a picture of the region with a slice taken parallel to the axis of rotation.

The slice is taken at a value of

The thickness of the slice and the shell is

The radius is

The height is

The shell has volume

The resulting solid has volume:

# = 2piint_0^2(28-4y-7y^2+y^3) dy#

# = 2pi[28y-2y^2 - 7/3y^3+y^4/4]_0^2#

#= (200pi)/3#

**If you prefer washers**

Then the thickness is

the greater radius is

the lesser radius is

# = pi int_1^5(14sqrt(x-1) - (x-1))dx#

To integrate I would substitute

See below.

#### Explanation:

Looking at the graph, we can see that the area **A** rotated around

We can find this volume by finding the volume of the area **(A+B)**. This is easy and doesnâ€™t require integration. The radius of the cylinder that will be formed has a **radius of 7**, this is just the height from the x axis. We square this and multiply by

Length of interval is

Volume (A + B ):

From this we need to subtract the volume of **B**. From the graph we can see that the radius is

Volume of B:

Area first:

..................................................................................................................................

Volume=

Volume of A:

Volume of revolution: