# Thread: Volume of an Ice Cream cone.

1. ## Volume of an Ice Cream cone.

I am quite perplexed by this problem and I have already graphed it within WinPlot. Would a good step be finding the plane of intersection?

"Find the volume of an ice cream cone bounded by the hemisphere
z = (8- x^2 - y^2)^(1/2) and the cone z = (x^2 + y^2) ^ (1/2)."

2. Originally Posted by EuptothePiI1
I am quite perplexed by this problem and I have already graphed it within WinPlot. Would a good step be finding the plane of intersection?

"Find the volume of an ice cream cone bounded by the hemisphere
z = (8- x^2 - y^2)^(1/2) and the cone z = (x^2 + y^2) ^ (1/2)."
Disclaimer I am using the spherical coordinates in the following form

$(\rho,\theta,\phi)$ where $\theta$ is the measure of the angle from the positive z axis and $\phi$ is the plane angle.
Alot of calculus books reverse the two angles.
First note that you will want to use spherical coordinates setting the equations equal gives

$8-x^2-y^2=x^2+y^2 \iff x^2+y^2=4$ plugging this into either equation gives $z=2$

So the circle of intersection of the two objects is described above

Using the above information we can find the angle from the positive z-axis to be

$\displaystyle \theta=\tan^{-2}\left( \frac{2}{2}\right)=\frac{\pi}{4}$

So the radius $\rho=\sqrt{8}$ and the plane angle $\phi$ needs to sweep out the whole circles so it goes from $0 \text{ to } 2\pi$

This gives

$\displaystyle \iiint_{V}dV=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{8}}\rho^2 \sin(\theta)\,d\rho \, d\theta \, d\phi$

3. I am going to attempt to follow what you showed me on paper now. I will edit this post once I come to a conclusion. Thank you very much for your time.

(15 minutes later, View Below)

Huzzah! Success! Thank you *very* much.