Expert: Paul Klarreich - 2/15/2010

Question
What is the volume of the solid obtained?

A region R is bounded by y=0, y=f(x)>0, x=o, x=t>0. If R is rotated around the x-axis, the volume of the resulting solid is (π/3)*(t^3+3t).

If the same region is rotated around the y-axis, what is the volume of the solid obtained?

I am completely stumped on this problem and was looking for some help in order to solve it. I am not sure if I should start with the given volume that is obtained when R is rotated around x-axis and work backwards to find out the volume when R is rotated around y-axis, OR if there is another way to go about finding the solution.

Answer
Questioner: Alex
Country: United States
Category: Calculus
Private: No
Subject: Calculus II
Question: What is the volume of the solid obtained?

A region R is bounded by y=0, y=f(x)>0, x=o, x=t>0. If R is rotated around the x-axis, the volume of the resulting solid is (π/3)*(t^3+3t).

If the same region is rotated around the y-axis, what is the volume of the solid obtained?

I am completely stumped on this problem and was looking for some help in order to solve it. I am not sure if I should start with the given volume that is obtained when R is rotated around x-axis and work backwards to find out the volume when R is rotated around y-axis, OR if there is another way to go about finding the solution.
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A region R is bounded by y=0, y=f(x)>0, x=o, x=t>0. If R is rotated around the x-axis, the volume of the resulting solid is (π/3)*(t^3+3t).

A V-of-R like this is integrated like this:

{ x=b
|    pi [f(x)]^2 dx
} x=a

Use the disk method -- each disk is  pi r^2 h, where your h = dx, and your  r = f(x)

In this case, your b = t, and your a = 0

{ t
|    pi [f(x)]^2 dx
} 0


Now if you integrate  [f(x)]^2 to  F(x) and then substitute:

pi F(x), from 0 to t,

you are supposed to get:

pi/3 (t^3 + 3t)

So it looks as if  F(t) = 1/3 (t^3 + 3t) = t^3/3 + t

That makes it look as if  F'(x) = x^2 + 1 = [f(x)]^2

Now you are on your way.  If y = f(x) = sqrt(x^2 + 1)

you can do your 'revolution about the y-axis', using

y = sqrt(x^2 + 1), or

y^2 = x^2 + 1,

x^2 = y^2 - 1

x = sqrt(y^2 - 1)

Now you can set up your second integral, using disks again. (Perhaps they will be rings, but no matter.)

Give it a try and let me know how it came out.