Calculus is not required. Matrices are helpful. Question is about volume of an ellipsoid.

This solution uses linear algebra . That was warned in the problem title with "matrices are helpful". Linear algebra is usually taught in most educational institutions after the full course of calculus. There is no requirement for this as the two subjects are quite separate. I encountered matrices in the form of Cramer's Rule in my seventh grade in the American public schools during my formal introduction to multiple equations in multiple unknowns .

Converting the polynomial to the matrix form used below (please see: augmented matrices and homogeneous coordinates ):

$\left(\begin{array}{cccc}x&y&z&1 \\ \end{array}\right) \left( \begin{array}{cccc} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{1,2} & a_{2,2} & a_{2,3} & a_{2,4} \\ a_{1,3} & a_{2,3} & a_{3,3} & a_{3,4} \\ a_{1,4} & a_{2,4} & a_{3,4} & a_{4,4} \\ \end{array} \right) \left(\begin{array}{c}x\\y\\z\\1 \\ \end{array}\right) \to \\ x^2 a_{1,1}+2 x y a_{1,2}+2 x z a_{1,3}+2 x a_{1,4}+y^2 a_{2,2}+2 y z a_{2,3}+2 y a_{2,4}+z^2 a_{3,3}+2 z a_{3,4}+a_{4,4}$

Equating the coefficients of the two polynomials and solving for the matrix elements gives the matrix below.

The surface equation can be stated as: $\left(\begin{array}{cccc}x&y&z&1 \\ \end{array}\right) \left( \begin{array}{cccc} 45472 & -16596 & -9600 & 16520 \\ -16596 & 55153 & -7200 & -72110 \\ -9600 & -7200 & 9225 & -3675 \\ 16520 & -72110 & -3675 & -469675 \\ \end{array} \right) \left(\begin{array}{c}x\\y\\z\\1 \\ \end{array}\right) = 0$

The center's coordinates are: $\left( \begin{array}{ccc} 45472 & -16596 & -9600 \\ -16596 & 55153 & -7200 \\ -9600 & -7200 & 9225 \\ \end{array} \right) \left(\begin{array}{c}x\\y\\z \\ \end{array}\right) = \left(\begin{array}{c}16520\\-72110\\-3675 \\ \end{array}\right)$ . $\left(\begin{array}{c}x\\y\\z \\ \end{array}\right) =\left( \begin{array}{ccc} \frac{12017}{285610000} & \frac{1461}{71402500} & \frac{128}{2142075} \\ \frac{1461}{71402500} & \frac{538}{17850625} & \frac{32}{714025} \\ \frac{128}{2142075} & \frac{32}{714025} & \frac{1321}{6426225} \\ \end{array} \right)\left(\begin{array}{c}16520\\-72110\\-3675 \\ \end{array}\right)=\left(\begin{array}{c}-1\\-2\\-3 \\ \end{array}\right)$ or the ellipsoid's center is at $x= 1,y= 2,z= 3$ .

The equation at the origin is $45472 x^2-33192 x y-19200 x z+55153 y^2-14400 y z+9225 z^2-608400$ . This was done by substituting the new definitions of the coordinates and simplifying. This translation of the origin does not change the volume.

The equation restated in the form used above is: $\left(\begin{array}{cccc}x&y&z&1 \\ \end{array}\right) \left( \begin{array}{cccc} 45472 & -16596 & -9600 & 0 \\ -16596 & 55153 & -7200 & 0 \\ -9600 & -7200 & 9225 & 0 \\ 0 & 0 & 0 & -608400 \\ \end{array} \right) \left(\begin{array}{c}x\\y\\z\\1 \\ \end{array}\right) = 0$ .

This is an interesting change in the central matrix's format and is characteristic of being at the origin.

The upper left 3x3 matrix's eigenvectors point in the directions to the ellipsoid's axes. If you do not understand that, then, stop, go read your linear algebra book to see what that means.

The eignvectors: $\left( \begin{array}{ccc} -3 & 4 & 0 \\ -48 & -36 & 25 \\ 4 & 3 & 12 \\ \end{array} \right)$

Because we want a rotation matrix, we will want to normalize the eigenvectors, that is, make their length equal to 1.

The rotation matrix: $\left( \begin{array}{ccc} -\frac{3}{5} & \frac{4}{5} & 0 \\ -\frac{48}{65} & -\frac{36}{65} & \frac{5}{13} \\ \frac{4}{13} & \frac{3}{13} & \frac{12}{13} \\ \end{array} \right)$

After the eignenvectors are normalized, taking the transpose of the result inverts the rotation matrix and makes it into the rotation matrix that puts the ellipsoid's axes onto the coordinate's axes. This transformation is also linear and does not change the ellipsoid's volume.

The inverse rotation matrix: $\left( \begin{array}{ccc} -\frac{3}{5} & -\frac{48}{65} & \frac{4}{13} \\ \frac{4}{5} & -\frac{36}{65} & \frac{3}{13} \\ 0 & \frac{5}{13} & \frac{12}{13} \\ \end{array} \right)$

To rotate the ellpsoid into the coordinate system axes: $x= -\frac{3 x}{5}-\frac{48 y}{65}+\frac{4 z}{13},y= \frac{4 x}{5}-\frac{36 y}{65}+\frac{3 z}{13},z= \frac{5 y}{13}+\frac{12 z}{13}$ .

This gives the equation for the centered, oriented ellipsoid of $16 x^2+9 y^2+z^2=144$ or $\left(\frac{x}{3}\right)^2+\left(\frac{y}{4}\right)^2+\left(\frac{z}{12}\right)^2=1$ .

Therefore, the volume divided by $\pi$ is $\frac43\times 3\times 4\times 12=192$ .

This methodology easily can be extended to two dimensions or even more dimensions.

The numbers were kept to relatively small integers. Pythagorean triangles were used to avoid algebraic numbers (radicials). Higher dimensions were not used. This was done to keep the problem easier to work.

Here is how the problem was derived:

The variables are $\{x,y,z\}$ .

The (center* is: $\{1,2,3\}$ .

The first rotation matrix for a rotation of $\tan ^{-1}(12,5)$ radians around the $x$ -axis based at the origin is: $\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \frac{12}{13} & -\frac{5}{13} & 0 \\ 0 & \frac{5}{13} & \frac{12}{13} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$ .

The second rotation matrix for a rotation of $\tan ^{-1}(3,4)$ radians around the $z$ -axis based at the origin is: $\left( \begin{array}{cccc} \frac{3}{5} & -\frac{4}{5} & 0 & 0 \\ \frac{4}{5} & \frac{3}{5} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$ .

The complete rotstion matrix was generated by: $\text{first}.\text{second}\to \left( \begin{array}{cccc} \frac{3}{5} & -\frac{4}{5} & 0 & 0 \\ \frac{48}{65} & \frac{36}{65} & -\frac{5}{13} & 0 \\ \frac{4}{13} & \frac{3}{13} & \frac{12}{13} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$

Write the centered, oriented form of the ellipsoid: $\left(\frac{x}{3}\right)^2+\left(\frac{y}{4}\right)^2+\left(\frac{z}{12}\right)^2=1$ .

The new variables in terms of the old variable after apply the combined rotation is $\left\{\frac{3 x}{5}-\frac{4 y}{5},\frac{48 x}{65}+\frac{36 y}{65}-\frac{5 z}{13},\frac{4 x}{13}+\frac{3 y}{13}+\frac{12 z}{13}\right\}$ .

The rotated polynomial: $\frac{1}{144} \left(\frac{4 x}{13}+\frac{3 y}{13}+\frac{12 z}{13}\right)^2+\frac{1}{16} \left(\frac{48 x}{65}+\frac{36 y}{65}-\frac{5 z}{13}\right)^2+\frac{1}{9} \left(\frac{3 x}{5}-\frac{4 y}{5}\right)^2-1=0 \to \frac{2842 x^2}{38025}-\frac{461 x y}{8450}-\frac{16 x z}{507}+\frac{55153 y^2}{608400}-\frac{4 y z}{169}+\frac{41 z^2}{2704}-1 = 0$ .

A second change of variables to move the ellipsoid from the origin: $x\to x-1,y\to y-2,z\to z-3$ gives: $\frac{1}{144} \left(\frac{4 (x-1)}{13}+\frac{3 (y-2)}{13}+\frac{12 (z-3)}{13}\right)^2+\frac{1}{16} \left(\frac{48 (x-1)}{65}+\frac{36 (y-2)}{65}-\frac{5 (z-3)}{13}\right)^2+\frac{1}{9} \left(\frac{3 (x-1)}{5}-\frac{4 (y-2)}{5}\right)^2-1=0$ ,

which simplifies to: $45472 x^2-33192 x y-19200 x z+33040 x+55153 y^2-14400 y z-144220 y+9225 z^2-7350 z-469675=0$ .

Then, I had to learn how to solve my own problem as I dislike putting up problems without solutions.

A few more hyperlinks from Wolfram Mathematica:

• Rotation matrix

• Eignvector

• Linear transformation

• Linear algebra topics