Integrating volume of a sphere with a cylinder "drilled" out of it

Question

Unfortunately, I am stuck again on another integration problem.

Famous last words, this should be simple.

$$ \text{A cylindrical drill with radius 5 is used to bore a hole through}\\\text{the center of a sphere of radius 7. Find the volume of the ring shaped solid that remains.} $$

So we can setup our problem by first defining our change in $\theta$ as the first region, because we can do some simple multiplication to fill the rest of the sphere symmetrically. We should just be able to calculate the change in $r$ inside of that.

$$ \begin{align} &=8 \int_{0}^{\frac{\pi}{4}} \int_{5}^{7} r\:dr\:d\theta\\ &=\frac{8}{2}\int_{0}^{\frac{\pi}{4}}49-25\:d\theta\\ &=24 \times \frac{8}{2} \times \frac{\pi}{4}\\ &=24\pi \end{align} $$

Edit: I attempted to re-evaluate my process, but the problem was still not correct.

I attempted to set my integrand to the arc length of the sphere - the arc length of the cylinder, but the integrand $\frac{r^3}{2}-5r$ was not the correct integrand to use.

Answer 1

Why not to use simple geometry, In general

The volume of the sphere, having a radius $R$, thoroughly drilled along its diameter by a cylinder having a radius $r$, is $$=\text{(volume of the original sphere)}-\text{(volume of of cylindrical hole with ends as spherical caps)}$$ $$V_{\text{drilled sphere}}=\frac{4\pi}{3}R^3-\frac{4\pi}{3}(R^3-(R^2-r^2)^{3/2})$$ $$\bbox[4pt, border:1px solid blue;]{\color{blue}{V_{\text{drilled sphere}}=\frac{4\pi}{3}(R^2-r^2)^{3/2}}} $$ where $\color{red}{0\leq r<R}$.

Now, substituting the given values, radius of the sphere, $R=7$ & the radius of the cylindrical drill $r=5$, we get the volume of ring shaped solid (i.e. drilled sphere) $$V_{\text{drilled sphere}}=\frac{4\pi}{3}(7^2-5^2)^{3/2}$$ $$=\frac{4\pi}{3}(24)^{3/2}$$ $$=\frac{4\pi}{3}(8)(6)^{3/2}$$ $$=\color{blue}{\frac{32\pi 6^{3/2}}{3}\approx 492.4991348\ldots \text{ cubic units}}$$

Answer 2

The integral you set up is giving you the area of 1/8 of an annulus with inner radius $5$ and outer radius $7$, not the volume of a solid.

Let's try finding the volume of the cylindrical hole and the volume of the sphere separately and then subtracting them.

For the hole, let's restrict ourselves to the first octant (as you did).

We can orient the solid so that the axis of the cylinder is the $z$-axis, and then, using cylindrical coordinates, we can find the limits of integration:

• $z$ goes from the $xy$-plane to the upper half of the sphere, so its limits are $0$ and $\sqrt{49-(x^2+y^2)}=\sqrt{49-r^2}$
• $r$ goes from $0$ to $5$
• $\theta$ goes from $0$ to $\frac \pi 2$.

Thus, the volume of $1/8$th of the hole is given by \begin{align}\int_0^{\pi/2}\int_0^5 \int_0^{\sqrt{49-r^2}}r\,dz\,dr\,d\theta&=\int_0^{\pi/2}\int_0^5 r\sqrt{49-r^2} \, dr \, d\theta \\ &=\int_0^{\pi/2} \frac{-1}{3}(49-r^2)^{3/2} \bigg|^5_0 \,d\theta \\ &=\int_0^{\pi/2} \frac{-1}{3}(24^{3/2}-343) \, d\theta \\ &=\frac{\pi}{6}(343-24^{3/2}) \end{align}

And subtracting (8 times) this from the volume of the sphere ($\frac{343\pi}{3}$ by geometry) should give you the volume of your solid.

Answer 3

This is the napkin ring problem.

This illustration hints at using Cavalieri's principle on it.

This section gives a complete solution.