Expert: Paul Klarreich - 3/22/2010

Question
The region bounded by y=sq.rt of x, y=0, x=0, and x=14 is revolved about the x-axis.  We need to find the value of x in the interval [0,14] that divides the solid into two parts of equal volume.  Then we need to find the value of x in the interval [0,14] that divides the solid into three parts of equal volume. Please show us all the steps in solving these problems.  

Thanks!

Answer
Questioner: Daphne
Country: United States
Category: Calculus
Private: No
Subject: Volume Under a curve
Question: The region bounded by y=sq.rt of x, y=0, x=0, and x=14 is revolved about the x-axis.  We need to find the value of x in the interval [0,14] that divides the solid into two parts of equal volume.  Then we need to find the value of x in the interval [0,14] that divides the solid into three parts of equal volume. Please show us all the steps in solving these problems.  

Thanks!
..................................
Here is your problem, rewritten.  

The region bounded by y=sqrt(x), y=0, x=0, and x = A is revolved about the x-axis.

Then its volume is given (using the disk method, of course -- my favorite.)

{A
|  pi y^2 dx
}0

{A
|  pi (sqrt(x))^2 dx
}0

{A
|  pi x dx = pi x^2/2, from 0 to A
}0

= pi A^2/2

And if A = 14, that is pi (14)^2/2 = 98pi

Now your question says:

For what value of A is pi A^2/2 equal to half of 98pi ?

And your second question says:

For what value of A is pi A^2/2 equal to one-third of 98pi ?

For what value of A is pi A^2/2 equal to two-thirds of 98pi ?

I think you can work it out now.