I am taking a Calculus course currently and am stuck on the last question of my assignment.

Find the volume of the region inside the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ and above the plane $z = b - y$.

I've tried solving the ellipsoid equation to get bounds on the triple integral, but it seems intractably hard, and I tried the change of coordinates $\beta = z + y$, $\gamma - (x+y)$, and $\lambda = x$. The Jacobian of this transformation has determinant 1 and should rotate $\mathbb{R}^3$ so that the plane $z=b-y$ is our "new" xy-plane. In any case, I haven't been able to figure it out using this approach either. Any help would be greatly appreciated!


Well, it is probably not the only way, but here it is: you can make a linear transformation of the space such that your ellipsoid becomes the unit sphere. This transformation will multiply all the volumes by $\frac{1}{abc}$. Next, plane $z = b - y$ will become some other plane. Then you'll simply need to calculate the volume of the shape bounded by the unit sphere and this new plane, and that cannot be too hard.

  • $\begingroup$ it wont let me comment, umm how do you transform the plane/whats the new plane? thanks $\endgroup$ – user140038 Apr 3 '14 at 7:30
  • $\begingroup$ @josh It is quite simple. The transformation $f: \mathbb{R}^3 \to \mathbb{R}^3$ that sends the original ellipsoid into a unit sphere is $f(x,y,z) = (x/a, y/b, z/c)$. The original plane is $P = \{(x,y,z) \mid z = b - y \}$. To find the transformed plane $f(P)$, note that $(x,y,z) \in f(P) \Leftrightarrow f^{-1}(x,y,z) \in P \Leftrightarrow (xa, yb, zc) \in P \Leftrightarrow zc = b - yb$. So the equation of the new plane is $zc = b - yb$. $\endgroup$ – Dan Shved Apr 3 '14 at 10:04
  • $\begingroup$ As a simple check of this answer, take some point on the original plane. For instance, $A = (x_0, y_0, z_0) = (0,0,b)$. Then find its image under $f$: $f(A) = f(x_0, y_0,z_0) = (0,0,b/c)$. Then check that $f(A)$ satisfies the equation that we have found: $(b/c)c = b - 0 \cdot b$. The equality is correct. You can also try some other points to be sure. For instance, take $B = (0,b,0)$. Clearly, $B \in P$. Now take $f(B) = (0,1,0)$. Does it belong to the new plane that we found? Yes, because $0 \cdot c = b - 1 \cdot b$ is true. $\endgroup$ – Dan Shved Apr 3 '14 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.