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Homework Statement
A sphere has a diameter of ##D = 2\rho = 4cm##. A cylindrical hole with a diameter of ##d = 2R = 2 cm## is bored through the center of the sphere. Calculate the volume of the remaining solid. (Spherical or cylindrical coordinates?)
hint: Place the shape into a convenient place in the xyz-coordinate system.
Homework Equations
Cylindrical coordinates:
\begin{cases}
x = rcos(\theta)\\
y = rsin(\theta)\\
z = z\\
x^2+y^2 = r^2\\
dV_c = rdzdrd\theta
\end{cases}
Spherical coordinates:
\begin{cases}
x = rcos(\theta) = \rho sin(\phi)cos(\theta)\\
y = rsin(\theta) = \rho sin(\phi)sin(\theta)\\
z = \rho cos(\phi)\\
x^2 + y^2 + z^2 = \rho^2\\
dV_s = \rho^2 sin(\phi)d\rho d\phi d\theta
\end{cases}
The Attempt at a Solution
I looked around the internet and found the exact same problem, except in that case the presenter of the question was kind enough to point out that this would be easier to do in cylindrical coordinates. In any case, my guess is that I should try to calculate the volume of the hole and subtract it from the volume of the sphere which we know to be ##V_s = \frac{4}{3}\pi \rho^3##. The hole seems to consist of a cylinder, with two lobes attached at each end.
Let us place the sphere's center at the origin (note that the small r in the pictures should be ##R## considering our derivation of the volume below):
If we first try cylindrical coordinates, we can easily calculate the volume of the cylinder:
\begin{align*}
V_c
&= \int_{0}^{2\pi}\int_{0}^{\pi}\int_{-\sqrt{\rho^2 - R^2}}^{\sqrt{\rho^2 - R^2}}rdzdrd\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\pi} 2R\sqrt{\rho^2 - R^2}drd\theta\\
&= \int_{0}^{2\pi} R^2\sqrt{\rho^2 - R^2}d\theta\\
&= 2 \pi R^2\sqrt{\rho^2 - R^2}
\end{align*}
As for the lobes, they are limited by
\begin{cases}
0 \leq \theta \leq 2\pi\\
0 \leq r \leq R\\
\sqrt{\rho^2 - R^2} \leq z \leq \sqrt{\rho^2 - r^2}
\end{cases}
Then the volume of our lobe is
\begin{align*}
V_L
&= \int_{0}^{2\pi}\int_{0}^{R}\int_{\sqrt{\rho^2 - R^2}}^{\sqrt{\rho^2 - r^2}} r dz dr d\theta \\
&= \int_{0}^{2\pi}\int_{0}^{R} r \left( \sqrt{\rho^2 - r^2}-\sqrt{\rho^2 - R^2} \right) dr d\theta\\
&= \int_{0}^{2\pi}\left(\int_{0}^{R} r\sqrt{\rho^2 - r^2} dr - \int_{0}^{R} r (\sqrt{\rho^2 - R^2} dr\right) d\theta\\
&= \int_{0}^{2\pi}
\left(\int_{0}^{R}\frac{-1}{2}\times \frac{2}{3}\times \frac{3}{2}\times -2r\sqrt{\rho^2 - r^2} dr
- \int_{0}^{R} r \stackrel{constant}{\sqrt{\rho^2 - R^2}} dr\right)
d\theta\\
&= \int_{0}^{2\pi} \left[ \frac{-2}{6}\sqrt{\rho^2 - r^2}^3\right]_0^R - \frac{1}{2}R^2\sqrt{\rho^2 - R^2} d\theta\\
&= \int_{0}^{2\pi} \frac{-1}{3}\sqrt{\rho^2 - r^2}^3 + \frac{1}{3}\rho^3 - \frac{1}{2}R^2\sqrt{\rho^2 - R^2}\ d\theta\\
&= \frac{-2\pi}{3}\sqrt{\rho^2 - r^2}^3 + \frac{2\pi}{3}\rho^3 - \frac{2\pi}{2}R^2\sqrt{\rho^2 - R^2}
\end{align*}
Now since in the case of our problem ##\rho = 2R##,
\begin{equation*}
\sqrt{\rho^2 - R^2} = \sqrt{(2R)^2 - R^2} = \sqrt{3R^2} = \sqrt{3}R
\end{equation*}
so
\begin{align*}
V_L
&= \frac{-2\pi}{3}\sqrt{\rho^2 - r^2}^3 + \frac{2\pi}{3}\rho^3 - \frac{2\pi}{2}R^2\sqrt{\rho^2 - R^2}\\
&= \frac{-2\pi}{3}(\sqrt{3}R)^3 + \frac{2\pi}{3}(2R)^3 - \pi R^2(\sqrt{3}R)\\
&= \pi R^3 \left( \frac{-6\sqrt{3}}{3} + \frac{16}{3} - \frac{3\sqrt{3}}{3}\right)\\
&=\frac{\pi R^3}{3}(16 - 9\sqrt{3})
\end{align*}
Then our volumes are:
\begin{cases}
Sphere: V_s &= \frac{4}{3}\pi \rho^3 \stackrel{\rho = 2R}{=} \frac{32}{3}\pi R^3\\
Cylinder: V_c &= 2\pi R^2 \sqrt{\rho^2 - R^2} = 2\sqrt{3}\pi R^3\\
Two \ lobes: 2V_L &= \frac{2\pi R^3}{3}(16 - 9\sqrt{3})
\end{cases}
and our total volume
\begin{align*}
V
&= V_s - V_c - 2V_L\\
&= \frac{32}{3}\pi R^3 - 2\sqrt{3}\pi R^3 - \frac{2\pi R^3}{3}(16 - 9\sqrt{3})\\
&= \frac{\pi R^3}{3} \left(32 - 2\sqrt{3} - 32 + 9\sqrt{3} \right)\\
&= \frac{12\sqrt{3}\pi R^3}{3}\\
&= 4\pi \sqrt{3} R^3\\
&\stackrel{R=1}{=} 4\pi \sqrt{3} \ (cm^3)
\end{align*}
which is the correct answer according to the question sheet.
Boom. Shaka. Laka.
I might get back to solving the spherical case in the future, but no way am I doing that now. This took me long enough with all the mistakes I made along the way and I have two more problems I need to get done by Monday.
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