# multiple integral where x,y,z are defined from R3 function.

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.

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?

$D$ 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 $512\pi/5$.
• I found a different solution to this without changing the coordinates. It's mostly based on algebra and it ends up like this:  \int Sep 8, 2016 at 9:10