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
Explanation:
If you need to deduce the volume of the ellipsoid here it is a way:
For a ellipsoid of such an equation:
We can find the volume by integration
-> 3D Variation: from
-> 2D Variation: from
-> 1D Variation: from
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#
