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
Dec 16, 2017

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 #y#, so we need to rewrite the curve #y=sqrt(x-1)# as #x = y^2+1#

enter image source here

The thickness of the slice and the shell is #dy#
The radius is #r = 7-y#
The height is #h = 5-(y^2+1) = 4-y^2#

The shell has volume #2pirhxx"thickness" = 2pi(7-y)(4-y^2)dy#

#y# varies from #0# to #2# (That is the #y# value on the curve at #x=5#.)

The resulting solid has volume:

#V = 2piint_0^2(7-y)(4-y^2)dy #

# = 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 #dx#,
the greater radius is #7#, and
the lesser radius is #sqrt(x-1)#.

#x# varies from #1# to #5#, so the volume of the solid is

#V = piint_1^5(7^2-(7-sqrt(x-1))^2)dx#

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

To integrate I would substitute #u = x-1# to get
#piint_0^4(14u^(1/2)-u)du#

Dec 16, 2017

See below.

Explanation:

enter image source here

Looking at the graph, we can see that the area A rotated around #y=7# is the volume we seek.

We can find this volume by finding the volume of the area (A+B). This is easy and doesn’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 #pi# x the length of the interval #[ 1,5]#

Length of interval is #5-1=4#

#:.#

Volume (A + B ):

#V=7^2pi*196pi#

From this we need to subtract the volume of B. From the graph we can see that the radius is #7-sqrt(x-1)#. This will have to be found by integration in the normal way.

Volume of B:

#(7-sqrt(x-1))^2=48-14sqrt(x-1)+x#

#V=pi*int_(1)^(5)(48-14sqrt(x-1)+x) dx=48x+1/2x^2-28/3(x-1)^(3/2)#

Area first:

#[48x+1/2x^2-28/3(x-1)^(3/2)]^(5)-[48x+1/2x^2-28/3(x-1)^(3/2)]_(1)#
..................................................................................................................................

#[48(5)+1/2(5)^2-28/3((5)-1)^(3/2)]^(5)-[48(1)+1/2(1)^2-28/3((1)-1)^(3/2)]_(1)#

#pi*[505/2-112/3sqrt(4)]-[97/2]=388/3pi#

Volume= #388/3pi#

Volume of A:

#196pi-388/3pi=200/3pi=66.667pi#

Volume of revolution:

enter image source here