# A Sphere of radius 2a has a hole of radius a drilled through the centre. What is the remaining volume?

Apr 5, 2017

$V = 4 \pi \sqrt{3} {a}^{3}$

#### Explanation:

Consider the section of the solid with a plane containing the axis of the cylinder and establish a system of reference with origin in the center of the sphere and with the axis of the cylinder as $x$ axis.

The volume of the remaining solid is generated by the rotation around this axis of the area delimited by the semi circle or radius $2 a$ and the line $y = a$.

For reasons of symmetry this is twice the volume of the area of the first quadrant comprised between the quarter circle of radius $2 a$, whose equation is $y = \sqrt{{\left(2 a\right)}^{2} - {x}^{2}}$ and the line $y = a$.

The two curves intercept when:

$a = \sqrt{{\left(2 a\right)}^{2} - {x}^{2}}$

${a}^{2} = 4 {a}^{2} - {x}^{2}$

$x = \sqrt{3} a$

The volume generated by the rotation of the area between $x$ and $x + \mathrm{dx}$ is then:

$\mathrm{dV} = \pi \left(4 {a}^{2} - {x}^{2} - {a}^{2}\right) = \pi \left(3 {a}^{2} - {x}^{2}\right)$

and integrating over the interval $\left(0 , \sqrt{3} a\right)$:

$V = 2 {\int}_{0}^{\sqrt{3} a} \pi \left(3 {a}^{2} - {x}^{2} \mathrm{dx}\right)$

$V = 2 \pi {\left[3 {a}^{2} x - {x}^{3} / 3\right]}_{0}^{\sqrt{3} a}$

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

We can also calculate this geometrically, since the volume of the remaining solid is the volume of the sphere ${V}_{s}$, minus the volume of the cylinder ${V}_{c}$, minus the volume of the two spherical caps ${V}_{p}$:

$V = {V}_{s} - {V}_{c} - 2 {V}_{p}$

Now:

${V}_{s} = \frac{4}{3} \pi {\left(2 a\right)}^{3} = \frac{32}{3} \pi {a}^{3}$

The height of the spherical caps is given by the formula:

$h = 2 a - \sqrt{{\left(2 a\right)}^{2} - {a}^{2}} = \left(2 - \sqrt{3}\right) a$

and their volume is:

${V}_{p} = \pi \frac{h}{6} \left(3 {a}^{2} + {h}^{2}\right) = \frac{\pi}{6} \left(2 - \sqrt{3}\right) a \left(3 {a}^{2} + {\left(\left(2 - \sqrt{3}\right) a\right)}^{2}\right) = \frac{\pi {a}^{3}}{6} \left(2 - \sqrt{3}\right) \left(3 + 4 - 4 \sqrt{3} + 3\right) = \frac{\pi {a}^{3}}{6} \left(2 - \sqrt{3}\right) \left(10 - 4 \sqrt{3}\right) = \frac{\pi {a}^{3}}{6} \left(32 - 18 \sqrt{3}\right)$

The heigth of the cylinder is the diameter of the sphere minus the height of the caps:

${h}_{c} = 4 a - 2 \left(2 - \sqrt{3}\right) a = a \left(4 - 4 + 2 \sqrt{3}\right) = 2 a \sqrt{3}$

so its volume is:

$V = \pi {a}^{2} {h}_{c} = 2 \pi {a}^{3} \sqrt{3}$

and finally:

$V = \frac{32}{3} \pi {a}^{3} - 2 \pi {a}^{3} \sqrt{3} - 2 \frac{\pi {a}^{3}}{6} \left(32 - 18 \sqrt{3}\right)$

$V = \pi {a}^{3} \left(\frac{32}{3} - 2 \sqrt{3} - \frac{1}{3} \left(32 - 18 \sqrt{3}\right)\right)$

$V = \pi {a}^{3} / 3 \left(32 - 6 \sqrt{3} - 32 + 18 \sqrt{3}\right)$

$V = \pi {a}^{3} / 3 \left(12 \sqrt{3}\right)$

$V = 4 \pi {a}^{3} \sqrt{3}$

Apr 5, 2017

$4 \sqrt{3} \pi {a}^{3}$

#### Explanation:

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system: The red circle has radius $2 a$, hence its equation is:

${x}^{2} + {y}^{2} = {\left(2 a\right)}^{2} \implies {x}^{2} + {y}^{2} = 4 {a}^{2}$

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius $2 a$. I will use the standard volume formula $V = \frac{4}{3} \pi {r}^{3}$

$\therefore {V}_{\text{sphere}} = \frac{4}{3} \pi {\left(2 a\right)}^{3}$
$\text{ } = \frac{4}{3} \pi 8 {a}^{3}$
$\text{ } = \frac{32}{3} \pi {a}^{3}$

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about $O y$, the shell formula is:

$V = 2 \pi \setminus {\int}_{\alpha}^{\beta} \setminus x f \left(x\right) \setminus \mathrm{dx}$

Also note that we require twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bore is given by:

${V}_{\text{bore}} = 2 \pi \setminus {\int}_{0}^{a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = 0 \\ x = a\end{matrix}\right. \implies \left\{\begin{matrix}u = 4 {a}^{2} \\ u = 3 {a}^{2}\end{matrix}\right.$

And so:

${V}_{\text{bore}} = - 2 \pi \setminus {\int}_{4 {a}^{2}}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{3 {a}^{2}}^{4 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{3 {a}^{2}}^{4 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{3 {a}^{2}}^{4 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(4 {a}^{2}\right)}^{\frac{3}{2}} - {\left(3 {a}^{2}\right)}^{\frac{3}{2}}\right)$
$\text{ } = \frac{4}{3} \pi \setminus \left(8 {a}^{3} - 3 \sqrt{3} {a}^{3}\right)$
$\text{ } = \frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}$

And so the total volume is given by:

 V_("total") = V_("sphere") - V_("bore"
$\text{ } = \frac{32}{3} \pi {a}^{3} - \left(\frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about $O y$, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bead is given by:

${V}_{\text{total}} = 2 \pi \setminus {\int}_{a}^{2 a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{a}^{2 a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = a \\ x = 2 a\end{matrix}\right. \implies \left\{\begin{matrix}u = 3 {a}^{2} \\ u = 0\end{matrix}\right.$

And so:

${V}_{\text{total}} = - 2 \pi \setminus {\int}_{3 {a}^{2}}^{0} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{3 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{0}^{3 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(3 {a}^{2}\right)}^{\frac{3}{2}} - 0\right)$
$\text{ } = \frac{4 \pi}{3} \setminus \left(3 \sqrt{3} {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$, as above