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?

1 Answer
Mar 4, 2016

992/3pi~=1038.820 " cubic units"

Explanation:

V_(Ellipsoid)=(4pi)/3*abc
Delta V=V_(Sphere)/2=(4pi)/3*r^3

V=V_0-Delta V
V=V_(Ellipsoid)-V_(Hemisphere)
V=(4pi)/3*6*7*12-(2pi)/3*8^3
V=672pi-1024/3pi=992/3pi~=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=-a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2) to x=a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2)
-> 2D Variation: from y=-b/c(c^2-z^2) to y=b/c(c^2-z^2)
-> 1D Variation: from z=-c to z=c

So

V=int_(-c)^c int_(-b/c(c^2-z^2))^(b/c(c^2-z^2)) int_( -a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2))^(a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2)) dx*dy*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*2*2*int_0^c int_0^(b/c(c^2-z^2)) int_0^(a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2)) dx*dy*dz
V=8*int_0^c int_0^(b/c(c^2-z^2)) x|_0^(a/(bc)*sqrt(b^2c^2-c^2y^2-b^2z^2)) dy*dz
V=(8a)/(bc)*int_0^c int_0^(b/c(c^2-z^2)) sqrt(b^2c^2-c^2y^2-b^2z^2) dy*dz
Making
cy=sqrt(b^2c^2-b^2z^2)sin alpha=b sqrt(c^2-z^2)sin alpha (remark that when y=0 then alpha=0, when y=b/c(c^2-z^2) then alpha =pi/2)
cdy=b sqrt(c^2-z^2)cos alpha dalpha => dy=b/csqrt(c^2-z^2)cos alpha dalpha

So the expression becomes

V=(8a)/(cancel(b)c)*int_0^c int_0^(pi/2) [cancel(b)sqrt(c^2-z^2)cos alpha][b/csqrt(c^2-z^2)cos alpha dalpha] dz
V=(8ab)/c^2*int_0^c int_0^(pi/2)(c^2-z^2)cos^2 alpha dalpha* dz
V=(8ab)/c^2*int_0^c int_0^(pi/2)(c^2-z^2)(cos (2alpha)/2+1/2)dalpha*dz
V=(8ab)/c^2*int_0^c (c^2-z^2)(sin (2alpha)/4+alpha/2)|_0^(pi/2)dz
V=(8ab)/c^2*(pi/4)int_0^c (c^2-z^2)dz
V=(2ab)/c^2*pi*(c^2z-z^3/3)|_0^c=(2ab)/c^2*pi*(c^3-c^3/3)=2abc*pi(2/3) => V_(Ellipsoid)=(4pi)/3abc