
Volume of ice cream cone
I need to find the volume of an ice cream cone bound by the hemisphere z = (8 = x^2  y^2)^1/2 and the cone z = (x^2 + y^2)^1/2
I know that is is a double integral but I can't even get started from there. Would someone please push or point me in the right direction? Thanks much for your effort. Frostking

$\displaystyle \sqrt{x^2+y^2}\leq z \leq \sqrt{8  x^2  y^2}$
The projection on the xyplane, $\displaystyle E$, is where the two surfaces meet: $\displaystyle \sqrt{x^2+y^2}=\sqrt{8  x^2  y^2} \Longleftrightarrow x^2+y^2=4$
So you can calculate it like this: $\displaystyle \iint_E \left( \sqrt{8  x^2  y^2}\sqrt{x^2+y^2}\right)dxdy$

To understand why we are doing this, consider this: $\displaystyle \int_{\sqrt{x^2+y^2}}^{\sqrt{8  x^2  y^2}}\left(\iint_E dxdy\right)dz=\iint_E \left( \sqrt{8  x^2  y^2}\sqrt{x^2+y^2}\right)dxdy$

So, if x^2 + y^2 is four then my bounds should be from 0 to 2 for both x and y is this correct? Frostking

$\displaystyle x^2+y^2=4$ is a circle with the radius 2. So both x and y go from 2 to 2. I always switch to polar coordinates for problems like these though.
Either go with $\displaystyle \sqrt{4x^2}\leq y \leq\sqrt{4x^2},\ 2\leq x\leq 2$ or the other way around $\displaystyle \sqrt{4y^2}\leq x \leq\sqrt{4y^2},\ 2\leq y\leq 2$