# volume of sphere using disk method.

• Mar 9th 2009, 04:43 PM
gammaman
volume of sphere using disk method.
Say I have a sphere and it has a radius of 3. How could I find it's volume using the disk method and integration?
• Mar 9th 2009, 06:40 PM
Mentia
By disk method I assume you mean using cylindrical coordinates. In this case you need to think about your limits on z. At what z is the top and bottom of the circle? At $\pm 3$.

So what is the equation of our circle? Its:

$x^2+y^2+z^2 = 9$

We know from polar coordinates that $r^2 = x^2 + y^2$. Then the equation of our circle is now:

$r^2+z^2 = 9$ or $r = \pm \sqrt[]{9-z^2}$
But in polar coordinates, a negative r doesn't make sense so we will say:
$r = \sqrt[]{9-z^2}$.

So now what is the area of any disk in our circle? It is:

$\int_{0 }^{2\pi } \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta$
But we want to add up all our disks, so we multiply them by a little bit of height, dz, and then integrate over z also:

$\int_{-3}^{3} \int_{0 }^{2\pi } \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta dz$

Does it work? Lets evaluate it, we know we should get $\frac{4}{3}\pi(3^3) = 36 \pi$

Indeed, if you went through and solved the triple integral you would get $36 \pi$.
• Mar 9th 2009, 06:45 PM
mollymcf2009
Quote:

Originally Posted by gammaman
Say I have a sphere and it has a radius of 3. How could I find it's volume using the disk method and integration?

Start by drawing the top half of a circle. Since I assume you were not given a specific location of the sphere on the graph, I would just center it at the origin. The top half of (since radius is 3) will go from (-3,0) to (3,0). Your disk will rotate about the x axis.

Since you are only using 1/2 of a circle, use the equation:

$y=\sqrt{r^2 - x^2}$

Volume $= 2\pi \int\limits^{3}_{-3} (\sqrt{9 - x^2})^2$

Can you take it from here?