I'm about to enter graduate school and I'm preparing for a placement exam involving some advanced calculus. I found this problem on one of the past exams and I've been stuck on it for awhile.
Find the volume of the solid that is bounded by the surface $$(x+y+z+5)^2+(x+2y+3z+6)^2+(x+3y+4z+7)^2=9$$
After plotting it in mathematica, I can see that this is a thin ellipsoid:
My intuition says to integrate over the elliptical cross sections, but I'm having issues setting this up. Some guidance would be much appreciated.
After expanding
$$ f(x,y,z) = f = (x + y + z + 5)^2 + (x + 2 y + 3 z + 6)^2 + (x + 3 y + 4 z + 7)^2 - 9 = 0 $$
we can arrange it as
$$ p^{\top}A p + B p + c = 0 $$
with $p = (x,y,z)^{\top}$ and
$$ A = \left( \begin{array}{ccc} 3 & 6 & 8 \\ 6 & 14 & 19 \\ 8 & 19 & 26 \\ \end{array} \right)\ \ \ B = (36,76,102) \ \ \ \mbox{and}\ \ \ c = 101 $$
Now calculating the $A$ eigenvectors we have
$$ T = \left( \begin{array}{ccc} 0.314338 & 0.733903 & 1. \\ -2.7732 & -0.174789 & 1. \\ 0.458862 & -1.55911 & 1. \\ \end{array} \right) $$
Making now the coordinate's change
$$ P = T p = (X,Y,Z) $$
we have in the new coordinates
$$ P^{\top}TAT^{\top}P+BT^{\top}P + c = 69.523 X^2+4.30315 Y^2+0.173815 Z^2+169.093 X-11.1191 Y+0.0263806 Z+101. $$
Now in this coordinate system we can calculate the ellipsoid three main axes $a,b,c$ and then the volume is the number
$$ V = \frac{4\pi}{3}a b c $$