# Using disk method to find a value of x

• July 26th 2010, 02:59 PM
Poptimus
Using disk method to find a value of x
Hey guys this problem has me stumped.

The region bounded by $y = \sqrt{x}$, $y = 0$, and $y = 4$ is revolved about the x-axis.

(a) Find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

(b) Find the values of x in the interval [0,4] that divide the solid into three parts of equal volume.

I suspect the easiest way would be to use the disk method; however, I'm not sure how to get an x value that would result in 2 or 3 parts of equal volume. I can guess and check for two, but there has to be an easier way.

Thanks
• July 26th 2010, 03:34 PM
skeeter
Quote:

Originally Posted by Poptimus
Hey guys this problem has me stumped.

The region bounded by $y = \sqrt{x}$, $y = 0$, and $y = 4$ is revolved about the x-axis.

(a) Find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

(b) Find the values of x in the interval [0,4] that divide the solid into three parts of equal volume.

I suspect the easiest way would be to use the disk method; however, I'm not sure how to get an x value that would result in 2 or 3 parts of equal volume. I can guess and check for two, but there has to be an easier way.

Thanks

first, find the overall volume, $V$, using the disk method from $x = 0$ to $x = 4$

then set up the equation ...

$\displaystyle \pi \int_0^a (\sqrt{x})^2 \, dx = \frac{V}{2}$

solve for $a$

set up the equations ...

$\displaystyle \pi \int_0^b (\sqrt{x})^2 \, dx = \frac{V}{3}$

$\displaystyle \pi \int_0^c (\sqrt{x})^2 \, dx = \frac{2V}{3}$

solve for $b$ and $c$
• July 26th 2010, 03:37 PM
drumist
Did you make a typo? Did you mean bound by $y=\sqrt{x}, ~ y=0, ~ x=4$ ?
• July 26th 2010, 03:41 PM
drumist
The inside of skeeter's integrals was not correct, but the important part was how he set up the endpoints. Just make sure you use the correct integrals though.
• July 26th 2010, 03:46 PM
Poptimus
Yeah that was a typo, it should have been $x = 4$. I'll see what I come up with, thanks guys.