Two approaches to the Volume of a Drilled Sphere (Self-Answered, open to other answers)

Question

Find the ratio of the volume of a sphere of Radius $R$ with a hole through its vertical axis in the shape of a coaxial cylinder with radius $r$, to the original volume of the the sphere with raidus $R$.

where $r = \dfrac{R}{6}$

The point of this is not to show exact equality, but approximations which change due to the size of $r$

I know to many this may just be a very simple problem, but its nice to see the connection between geometric and algebraic scenarios once more.

Any other interesting solutions to this would be nice to see!

Answer 1

We assume that $r<R$. The volume of the "cylinder", is $$ V_{\text{cyl}}=\iiint_{\text{cyl}}1\,dx\,dy\,dz=\iint_{D} \Bigl(\sqrt{R^2-x^2-y^2}-\bigl(-\sqrt{R^2-x^2-y^2}\bigr)\Bigr)\,dx\,dy, $$ where $D=\{(x,y)~|~x^2+y^2\leq r^2\}$. Here we used that the cylinder is bounded below by $z=-\sqrt{R^2-x^2-y^2}$ and above by $z=\sqrt{R^2-x^2-y^2}$. Switching to polar coordinates, this becomes $$ V_{\text{cyl}}=\iint_{E}2\sqrt{R^2-\rho^2}\rho\,d\rho d\theta, $$ where $E=\{(\rho,\theta)~|~0\leq\rho\leq r,\ 0\leq\theta\leq2\pi\}$. Iterating, we find that $$ \begin{aligned} V_{\text{cyl}}&=\int_0^{2\pi}1\,d\theta \int_0^r2\sqrt{R^2-\rho^2}\rho\,d\rho=2\pi\cdot2\cdot\Bigl[-\frac{1}{3}(R^2-\rho^2)^{3/2}\Bigr]_{\rho=0}^r\\ &=\frac{4\pi}{3}\Bigl(R^3-\bigl(R^2-r^2\bigr)^{3/2}\Bigr) \end{aligned} $$ I assume that the volume $V_{\text{ball}}=4\pi R^3/3$ of the ball is known, and thus, the volume of the ball minus the cylinder is $$ V=V_{\text{ball}}-V_{\text{cyl}}=\frac{4\pi}{3}\bigl(R^2-r^2\bigr)^{3/2}. $$

Update after the question was updated

The ratio of $V/V_{\text{ball}}$ is therefore $$ \frac{V}{V_{\text{ball}}}=\frac{\bigl(R^2-r^2\bigr)^{3/2}}{R^3} $$ which equals $$ \frac{35\sqrt{35}}{216} $$ if $r=R/6$.

Answer 2

I found out that there is a simpler geometric way of solving this, and a calculus method. I shall demonstrate both:

1. Geometric Approach:

Start off by determining the the volume of the original sphere (without the hole), is given by the formula for the volume of the sphere:

$$\Large{V_S=\dfrac{4}{3}\pi R^3\quad\quad \text{ where R = Sphere radius}}$$

Now we need to find the volume of the drilled sphere. You need to realize that the volume of the drilled sphere must be: $$\text{(Volume of original sphere)}-\text{(Volume of empty cylinder)}$$

The equation for the volume of a cylinder is given by:

$$V_C = \pi r^{2} h\quad\quad\text{ r = Cylinder radius, h = Cylinder height} $$

Substituting in $r = \dfrac{R}{6}$ and $h = 2R$:

$$\large{V_C=\pi \left(\dfrac{R}{6}\right)^2 (2R)=\dfrac{\pi R^3}{18}}$$

Thus, the volume of the drilled sphere must be:

$$\Large{V_{drill}=V_S-V_C=\dfrac{4\pi R^3}{3}-\dfrac{\pi R^3}{18} = \dfrac{23\pi R^3}{18} }$$

The ratio of the volume of the drilled sphere to the volume of the original sphere is then given by:

$$\large{\dfrac{\text{(Volume of drilled sphere)} }{\text{(Volume of original sphere)} } = \dfrac{\dfrac{23\pi R^3}{18} }{\dfrac{4\pi R^3}{3}} = \dfrac{23}{24} \approx \boxed{0.959} }$$

2. Calculus Approach:

To do this, we will utilize the power of the washer method, which defines the volume of rotation about an axis defined as a difference of two functions is:

$$\Large{V = \pi \int_a^b (\text{Outer Radius} )^2 - (\text{Inner Radius})^2 \, \mathrm{d}x}$$

If we are to take a vertical slice of the sphere in the 1st quadrant, we recieve the equation of the quarter circle defined by the slice as:

$$\Large{ f(x) = +\sqrt{R^2-x^2} }$$

where R is the radius of the sphere, and we take the positive root as we are in the first quadrant. We should also notice that with this method, the slice of the cylinder will be defined as the vertical line $ x= r$ where r is the cylinder's radius.

Since are trying to find horizontal disks that are created by revolving our functions around the y-axis, we will need to find inverse so that y is our variable, because the rectangles formed to construct the disk lie parallel to the x-axis. (Comment if that doesn't make too much sense; a diagram would probably be good, but I don't know how to make one online)

Luckily for us, the equation of the cylinder is always a constant function of y. Luckily once more, the inverse of our quarter circle is also very simple:

$$\Large{f(y) = +\sqrt{R^2-y^2} \\ g(y) = r = \dfrac{R}{6}}$$

When integrating with respect to the y-axis, you must subtract the function closer to the y-axis from the one that is farther. As we can see in the given picture, the sphere is on the "right" side of the cylinder, up until the point $x = r $ or $f(r)=\sqrt{R^2-r^2}=\sqrt{R^2-\dfrac{R^2}{36}} = \sqrt{\dfrac{35R^2}{36}} = \dfrac{R\sqrt{35}}{6}$, where the hole begins to touch the sphere. The bounds when integrating with respect to y are from the smallest y value attained to the largest. Putting this all together we get:

$$\large{V=\pi \int_0^{\frac{R\sqrt{35}}{6}} (f(y))^2-(g(y))^2 \, dy =\pi \int_0^{\frac{R\sqrt{35}}{6}} \left(\sqrt{R^2-y^2}\right)^2-\left(\dfrac{R}{6}\right)^2 \, dy \\ = \pi \int_0^{\frac{R\sqrt{35}}{6}} R^2-y^2-\dfrac{R^2}{36} \,dy =\pi \int_0^{\frac{R\sqrt{35}}{6}} \dfrac{35R^2}{36}-y^2\,dy = \pi\left. (\dfrac{35yR^2}{36}-\dfrac{y^3}{3})\right\vert_0^{\frac{R\sqrt{35}}{6}}\\ \pi\left(\dfrac{35\sqrt{35}R^3}{216} - \dfrac{35\sqrt{35}R^3}{648}\right) = \dfrac{35\pi\sqrt{35} R^3}{324}}$$

This is the volume of half of the drilled sphere as we have only rotated formed a figure in the upper quadrants. Thus the ratio would be the same whether we double this value, or half the value of the original sphere. We choose the latter:

$$\large{\dfrac{\text{(Half of Volume of drilled sphere)} }{\text{(Half of Volume of original sphere)} } = \dfrac{\dfrac{35\pi\sqrt{35} R^3}{324}}{\dfrac{2\pi R^3}{3}} = \dfrac{35\sqrt{35}}{216} \approx \boxed{0.959} }$$