# Thread: Using disk method to find a value of x

1. ## 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

2. 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 $c$

3. Did you make a typo? Did you mean bound by $y=\sqrt{x}, ~ y=0, ~ x=4$ ?

4. 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.

5. Yeah that was a typo, it should have been $x = 4$. I'll see what I come up with, thanks guys.