# Finding the volume between a cone and a sphere

I have to find the volume between the sphere $x^2+y^2+z^2=1$ and below the cone $z=\sqrt{x^2+y^2}$ using Spherical Coordinates.

Here is what I have so far:

Transforming the cone part gives:

\begin{align*} \sqrt{x^2+y^2} &= \sqrt{r^2\cos{\theta}^2\sin{\phi}^2+r^2\sin{\theta}^2\sin{\phi}^2}\\ &=\sqrt{r^2\sin{\phi}^2(\cos{\theta^2}+\sin{\theta}^2)}\\ &=r\sin{\phi} \end{align*}

I know the shape of intersection is a circle with $r = \frac{1}{4}$.

I am pretty sure that as far as the limits of integration will go, $0\leq\theta\leq 2\pi$, but I have no idea where to go from here.

How can I set up the integral?

• Dec 14, 2014 at 16:55
• @user_of_math That question is very much related, but I have no idea what is supposed to go inside the integral for example, and the answer and question don't help me out there Dec 14, 2014 at 16:56
• Te intersection of sphere with the cone is when $r \sin \phi = \frac{1}{\sqrt{2}}$ Nov 4, 2015 at 3:00

I'm adding this answer because the other answers so far answer different questions (either a different region, or a different method) than what was originally asked. I'm also going to try to explain my work in more depth to make it easier to follow along.

Note that the cone $$z = \sqrt{(x^2 + y^2)}$$ can be represented in spherical coordinates as $$\phi = \frac{\pi}{4}$$. You already calculated that $$z = r \sin \phi$$, and we know from the spherical coordinate transform that $$z = r \cos \phi$$. Combining these two equations, we get $$\sin \phi = \cos \phi$$ or $$tan \phi = 1$$, which means $$\phi = \frac{\pi}{4}$$. Intuitively/geometrically, you can also imagine that the cone covers the full range from $$\theta = 0$$ to $$\theta = 2 \pi$$ of rotation around the z-axis, meaning $$\theta$$ can be any value, and the radius can be any value as well (the cone is unbounded above), so the only parameter needed to define the cone is $$\phi$$.

To set up the region of integration, we know that to calculate the volume of the sphere, we would have the radius ranging from 0 to 1, with $$\theta$$, the angle of rotation around the z-axis, ranging from 0 to $$2 \pi$$, and $$\phi$$, the angle with the z-axis, ranging from 0 to $$\pi$$. In this case however, we want to exclude the region of the cone, which is the region from $$\phi = 0$$ to $$\phi = \frac{\pi}{4}$$. Thus we want to include $$\phi$$ only over the range from $$\frac{\pi}{4}$$ to $$\pi$$.

Thus, we have the below integral:

$$\int_{0}^{1} \int_{0}^{2 \pi} \int_{\frac{\pi}{4}}^{\pi} r^2 sin(\phi) d\phi d\theta dr$$

where $$r^2 sin(\phi)$$ comes from the change of base formula, and can be calculated as the absolute value of the determinant of the derivative matrix of $$x$$, $$y$$, $$z$$ with respect to $$r$$, $$\theta$$, and $$\phi$$.

The integral can be evaluated in any order, probably $$\theta$$ then $$r$$ then $$\phi$$ to get $$2 \pi \frac{1}{3} (1 + \frac{1}{\sqrt{2}}) \approx 3.58$$.

A quick intuitive check confirms that this answer makes sense: the volume of the sphere is given by $$\frac{4}{3}\pi r^3$$ which in this case is just $$\frac{4}{3}\pi \approx 4.19$$, and we're taking out a small fraction of the top half of the sphere.

The most advisable is to use the Change of Variables Theorem, in that sense, we have the integral of the region is:

$$\int_{0}^{\frac{1}{\sqrt{2}}}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}r^{2}\sin \left(\phi \right) d\phi d\theta dr=\frac{1}{6}\left(\sqrt{2}-1\right)\pi$$ The calculations of the integral is left as an exercise.

• Where do you get the 1/sqrt(2) from for the most outer integral? Apr 1, 2021 at 11:17

HINT:

This is not a proper answer using spherical coordinates so giving as hint for answer, hope details are easy to fill in. Area of spherical cap is area of equatorial circle multiplied by cap height.

$$\frac{Area_{spherical cap}}{ 4 \pi R^2} \cdot (\frac43 \pi R^3)$$

$$= 2\pi R^2 (1- \frac{1}{\sqrt2})\cdot \frac{R}{3}$$

Maximum radius of cone is $\frac{R}{\sqrt2}.$