It's an exercise i'm stuck with for some time.

We have integral

$$\iiint_D 4x^2 dx dy dz,$$ where $x,y,z$ belong to $D=\{ (x,y,z) \in \mathbb R^3 : (x^2/4) + (y^2/9) + (z^2/16) \le 1\}$.

I know how to solve the integral but i cant define $x,y,z$

Any help with D will be great.

Adittion 9/8/2016 :

I found a theorem saying that:

$$\iiint_D f\ dx dy dz = \int_a^b \int_h^h \int_g^g f(x,y,z) dzdydx $$

if D={xe[x0,x1] , ye[h0(x),h1(x) , ze[g0(x,y), g1{x,y) }

here is the link: http://users.math.msu.edu/users/gnagy/teaching/10-fall/mth234/w10-234-h.pdf it is in page 1

I tried this solution and found it extremely difficult to evaluate the integral to the end. Is the use of this theorem correct in this case? Is there an easiest way to solve this?


1 Answer 1


$D$ is an ellipsoid and you can easily compute your integral by a spherical change of coordinates: $x=2\rho \cos\vartheta_1$, $y=3\rho \sin\vartheta_1\cos\vartheta_2$, $z=4\rho\sin\vartheta_1\sin\vartheta_2$. In these variables $$ D=\{(\rho,\vartheta_1,\vartheta_2)\in\mathbb R^3\mid 0\leqslant\rho\leqslant 1\,,0\leqslant\vartheta_1\leqslant\pi\,,0\leqslant\vartheta_2\leqslant 2\pi\}$$ whereas the integral becomes

$$\int_D384\rho^4\cos^2\vartheta_1\sin\vartheta_1\, d\rho\, d\vartheta_1\,d\vartheta_2 = 384\cdot\int_0^1\rho^4\, d\rho\cdot\int_0^\pi \cos^2\vartheta_1\sin\vartheta_1\, d\vartheta_1\cdot\int_0^{2\pi}d\vartheta_2 $$

Now I let you conclude the computation! The result (edited on sept 9) is $512\pi/5$.

  • $\begingroup$ Can you explain what you did there please? I'm not familiar with this. $\endgroup$
    – George Sp
    Sep 7, 2016 at 17:24
  • $\begingroup$ My suggestion is about the change of variables in multiple integrals. You should learn about it in order to solve this kind of exercises. $\endgroup$ Sep 7, 2016 at 19:55
  • $\begingroup$ I found a different solution to this without changing the coordinates. It's mostly based on algebra and it ends up like this: $$ \int $\endgroup$
    – George Sp
    Sep 8, 2016 at 9:10
  • $\begingroup$ Using the theorem you cited is correct too, but somehow longer. $\endgroup$ Sep 8, 2016 at 12:28
  • $\begingroup$ Thanks for the help @james. Can you please tell me what did you use to decide how to change the coordinates? Is it based on some algorithm/theorem or it's always the same for ellipsoids' ? $\endgroup$
    – George Sp
    Sep 8, 2016 at 16:02

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