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?
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 $\displaystyle \pm 3$.
So what is the equation of our circle? Its:
$\displaystyle x^2+y^2+z^2 = 9$
We know from polar coordinates that $\displaystyle r^2 = x^2 + y^2$. Then the equation of our circle is now:
$\displaystyle r^2+z^2 = 9$ or $\displaystyle r = \pm \sqrt[]{9-z^2}$
But in polar coordinates, a negative r doesn't make sense so we will say:
$\displaystyle r = \sqrt[]{9-z^2}$.
So now what is the area of any disk in our circle? It is:
$\displaystyle \int_{0 }^{2\pi } \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta$
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:
$\displaystyle \int_{-3}^{3} \int_{0 }^{2\pi } \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta dz$
Does it work? Lets evaluate it, we know we should get $\displaystyle \frac{4}{3}\pi(3^3) = 36 \pi$
Indeed, if you went through and solved the triple integral you would get $\displaystyle 36 \pi$.
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:
$\displaystyle y=\sqrt{r^2 - x^2}$
Volume $\displaystyle = 2\pi \int\limits^{3}_{-3} (\sqrt{9 - x^2})^2$
Can you take it from here?