An ellipsoid has radii with lengths of 6 , 7 , and 12 . A portion the size of a hemisphere with a radius of 8  is removed form the ellipsoid. What is the remaining volume of the ellipsoid?

Mar 4, 2016

$\frac{992}{3} \pi \cong 1038.820 \text{ cubic units}$

Explanation:

${V}_{E l l i p s o i d} = \frac{4 \pi}{3} \cdot a b c$
$\Delta V = {V}_{S p h e r e} / 2 = \frac{4 \pi}{3} \cdot {r}^{3}$

$V = {V}_{0} - \Delta V$
$V = {V}_{E l l i p s o i d} - {V}_{H e m i s p h e r e}$
$V = \frac{4 \pi}{3} \cdot 6 \cdot 7 \cdot 12 - \frac{2 \pi}{3} \cdot {8}^{3}$
$V = 672 \pi - \frac{1024}{3} \pi = \frac{992}{3} \pi \cong 1038.820$

If you need to deduce the volume of the ellipsoid here it is a way:

For a ellipsoid of such an equation:
${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} + {z}^{2} / {c}^{2} = 1$
We can find the volume by integration
-> 3D Variation: from $x = - \frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}$ to $x = \frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}$
-> 2D Variation: from $y = - \frac{b}{c} \left({c}^{2} - {z}^{2}\right)$ to $y = \frac{b}{c} \left({c}^{2} - {z}^{2}\right)$
-> 1D Variation: from $z = - c$ to $z = c$

So

$V = {\int}_{- c}^{c} {\int}_{- \frac{b}{c} \left({c}^{2} - {z}^{2}\right)}^{\frac{b}{c} \left({c}^{2} - {z}^{2}\right)} {\int}_{- \frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}}^{\frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}} \mathrm{dx} \cdot \mathrm{dy} \cdot \mathrm{dz}$

Since the partial volumes of the 8 spatial regions (limited by the planes formed by the x-axis, y-axis and z-axis) are equal:

$V = 2 \cdot 2 \cdot 2 \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{b}{c} \left({c}^{2} - {z}^{2}\right)} {\int}_{0}^{\frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}} \mathrm{dx} \cdot \mathrm{dy} \cdot \mathrm{dz}$
$V = 8 \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{b}{c} \left({c}^{2} - {z}^{2}\right)} x {|}_{0}^{\frac{a}{b c} \cdot \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}}} \mathrm{dy} \cdot \mathrm{dz}$
$V = \frac{8 a}{b c} \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{b}{c} \left({c}^{2} - {z}^{2}\right)} \sqrt{{b}^{2} {c}^{2} - {c}^{2} {y}^{2} - {b}^{2} {z}^{2}} \mathrm{dy} \cdot \mathrm{dz}$
Making
$c y = \sqrt{{b}^{2} {c}^{2} - {b}^{2} {z}^{2}} \sin \alpha = b \sqrt{{c}^{2} - {z}^{2}} \sin \alpha$ (remark that when $y = 0$ then $\alpha = 0$, when $y = \frac{b}{c} \left({c}^{2} - {z}^{2}\right)$ then alpha =pi/2)
$c \mathrm{dy} = b \sqrt{{c}^{2} - {z}^{2}} \cos \alpha \mathrm{da} l p h a$ => $\mathrm{dy} = \frac{b}{c} \sqrt{{c}^{2} - {z}^{2}} \cos \alpha \mathrm{da} l p h a$

So the expression becomes

$V = \frac{8 a}{\cancel{b} c} \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{\pi}{2}} \left[\cancel{b} \sqrt{{c}^{2} - {z}^{2}} \cos \alpha\right] \left[\frac{b}{c} \sqrt{{c}^{2} - {z}^{2}} \cos \alpha \mathrm{da} l p h a\right] \mathrm{dz}$
$V = \frac{8 a b}{c} ^ 2 \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{\pi}{2}} \left({c}^{2} - {z}^{2}\right) {\cos}^{2} \alpha \mathrm{da} l p h a \cdot \mathrm{dz}$
$V = \frac{8 a b}{c} ^ 2 \cdot {\int}_{0}^{c} {\int}_{0}^{\frac{\pi}{2}} \left({c}^{2} - {z}^{2}\right) \left(\cos \frac{2 \alpha}{2} + \frac{1}{2}\right) \mathrm{da} l p h a \cdot \mathrm{dz}$
$V = \frac{8 a b}{c} ^ 2 \cdot {\int}_{0}^{c} \left({c}^{2} - {z}^{2}\right) \left(\sin \frac{2 \alpha}{4} + \frac{\alpha}{2}\right) {|}_{0}^{\frac{\pi}{2}} \mathrm{dz}$
$V = \frac{8 a b}{c} ^ 2 \cdot \left(\frac{\pi}{4}\right) {\int}_{0}^{c} \left({c}^{2} - {z}^{2}\right) \mathrm{dz}$
$V = \frac{2 a b}{c} ^ 2 \cdot \pi \cdot \left({c}^{2} z - {z}^{3} / 3\right) {|}_{0}^{c} = \frac{2 a b}{c} ^ 2 \cdot \pi \cdot \left({c}^{3} - {c}^{3} / 3\right) = 2 a b c \cdot \pi \left(\frac{2}{3}\right)$ => ${V}_{E l l i p s o i d} = \frac{4 \pi}{3} a b c$