Question

Cylindrical hole of radius a is bored through a sphere of radius 2a. the surface of the hole passes through the center of the sphere. how much material is removed

Answer 1

The amount of material removed is the volume of the region within the sphere bounded by the cylinder. Consider a sphere of radius $2a$ centered at the origin; this sphere has equation

$x^2+y^2+z^2=4a^2\iff z^2=4a^2-x^2-y^2$

The given cylinder has equation

$x^2+y^2=a^2$

The volume of the region of interest $\mathcal D$ is given by

$\displaystyle\iiint_{\mathcal D}\mathrm dV$

Converting to cylindrical coordinates, setting

$x=r\cos\theta$

$y=r\sin\theta$

$z=\zeta$

we have

$z^2=4a^2-r^2$

and

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\left|\dfrac{\partial(x,y,z)}{\partial(r,\theta,\zeta)}\right|\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$

$\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$

where $\dfrac{\partial(x,y,z)}{\partial(r,\theta,\zeta)}$ is the Jacobian of the transformation from $(x,y,z)$ to $(r,\theta,\zeta)$. The region $\mathcal D$ is described by the set

$\left\{(r,\theta,\zeta)\,:\,0\le r\le a\land0\le\theta\le2\pi\land-\sqrt{4a^2-r^2}\le\zeta\le\sqrt{4a^2-r^2}\right\}$

The integral is then

$\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=a}\int_{\zeta=-\sqrt{4a^2-r^2}}^{\zeta=\sqrt{4a^2-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

The integral with respect to $\zeta$ is symmetric about $\zeta=0$, so we instead compute twice the integral from $\zeta=0$ to $\zeta=\sqrt{4a^2-r^2}$, and we can immediately compute the integral with respect to $\theta$:

$=\displaystyle4\pi\int_{r=0}^{r=a}r\sqrt{4a^2-r^2}\,\mathrm dr$

Now, let $s=4a^2-r^2$, so that $\mathrm ds=-2r\,\mathrm dr$:

$=\displaystyle-2\pi\int_{r=0}^{r=a}-2r\sqrt{4a^2-r^2}\,\mathrm dr=-2\pi\int_{s=4a^2}^{s=3a^2}\sqrt s\,\mathrm ds$

$=-2\pi\cdot\dfrac23s^{3/2}\bigg|_{s=4a^2}^{s=3a^2}$

$=\dfrac{4\pi}3\left((4a^2)^{3/2}-(3a^2)^{3/2}\right)$

$=\dfrac{4\pi(8-3^{3/2})a^3}3$