# Find the mass of the ice cream cone

The region looks like an ice cream cone. It is an upside-down circular cone attached to a slice of a sphere. I'm pretty sure the way you are "supposed" to solve it is with spherical coordinates, since all of the numbers become suspiciously convenient.

I know the formula for mass is $$\int \int \int\,\, [\text{density}] \,\,dV$$

I will do $$dr\, d\phi \,d\theta$$. I think $$r$$ goes between $$0$$ and $$2\sqrt{2}$$, and $$\phi$$ goes between $$0$$ and $$\pi/4$$ (is the angle of the cone $$45^\circ$$?), and $$\theta$$ goes between $$0$$ and $$2\pi$$.

Also, $$x = r\sin\phi \cos \theta$$ $$y = r\sin\phi \sin \theta$$ $$z = r\cos\phi$$

And that means that the density, which is $$x^2 + y^2 + z^2$$, is just equal to $$r^2$$. We also have to multiply by $$r^2 \sin \phi$$ because it's spherical coordinates.

So the mass is

$$\text{mass} = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sqrt{2}} \,r^2 \left(r^2 \sin \phi\right)\,\,dr\, d\phi \,d\theta$$

I got $$\frac{256 \pi \sqrt{2} - 256 \pi}{5}$$ but that isn't one of the answers.

What did I do wrong? Or is the question wrong?

• See nothing wrong with your working. Oct 18, 2020 at 21:33

The angle of $$\phi$$ can by determined by substituting function
$$x^2+y^2=z^2 \to (using spherical coordinates) \to r^2sin^2(\phi)cos^2(\theta)+r^2sin^2(\phi)sin^2(\theta)=r^2cos^2(\phi)$$ $$\to sin^2(\phi) = \cos^2(\phi)$$ This condition is met for $$\phi = \pi/4$$ so there is your angle.
I calculated the mass to be $$2^6\pi\frac{1}{5}\sqrt{2}^5(1-\frac{\sqrt{2}}{2})$$
Hopefully it will be helpful. At least your question about the angle being $$\pi/4$$ is answered.