# How to find the bounds of the volume integral $\int_\Omega (6xz + 2y +3z^2) \, \text{d} \tilde{x}$?

I'm studying on integrating over volumes and I don't know how to set the bounds in this exercise:

Let $$\Omega := \left\{ (x,y,z) \in \mathbb{R}^3 \,\big| \,\frac{x^2}{4} + y^2 + \frac{z^2}{9} <1 \right\}$$ and $$\tilde{x} = (x,y,z)$$.

I want to evaluate the following integral

\begin{align} \int_\Omega (6xz + 2y +3z^2) \ \text{d} \tilde{x}. \end{align}

I reckon it's best to start off with the first integrand, using Fubini:

\begin{align} \int_\Omega 6xz \ \text{d} \tilde{x} = 6 \int_\Omega xz \ \text{d}x \text{d}y \text{d}z \\ 6 \int_?^? z \int_?^? \int_?^? x \ \text{d}x \text{d}y \text{d}z, \end{align}

but I don't know how to define the bounds for each of the integrals. Obviously it depends on $$\Omega$$, and its easy to see that $$\Omega$$ defines a contorted 3D-ellipsoid. My intuition is somehow using sphere coordinates, but Im just not sure about the exact procedure at all on how exactly can I set the bounds, any ideas?

Thanks

EDIT: Thanks to @heropup's answer, I can edit my question by the additional knowledge I have now. Mind that Im a newbie, so please point out anything that has room for improvement.

First, we can transform $$\Omega$$ into a unit 3D-sphere, by using the transformation $$(x,y,z) \mapsto (2u,v,3w)$$, which gives us

\begin{align} \tilde{\Omega} := \left\{ (2u,v,3w) \in \mathbb{R}^3 \,\Big| \,\frac{(2u)^2}{4} + v^2 + \frac{(3w)^2}{9} = u^2 + v^2 + w^2 < 1\right\}. \end{align}

The corresponding integral (using Transformation theorem) turns into

\begin{align} \int_{\tilde{\Omega}} (36uw + 2v + 27w^2) \det \frac{\partial (x,y,z)}{\partial(u,v,w)} \ \ \text{d} u \text{d} v \text{d} w, \end{align}

where

\begin{align} \det \frac{\partial (x,y,z)}{\partial(u,v,w)} = \det \begin{pmatrix} 2 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = 6, \end{align}

so

\begin{align} \int_{\tilde{\Omega}} (216uw + 12v + 162w^2) \ \text{d} u \text{d} v \text{d} w. \end{align}

We can now solve this by using spherical coordinates

\begin{align} u = r \, \sin(\phi) \cos(\theta) \\ v = r \, \sin (\phi) \sin(\theta) \\ w = r \, \cos(\phi) \end{align}

and the Jacobian determinate $$r^2 \sin(\phi)$$ to derive the following integral

\begin{align} \int_0^{2\pi} \int_0^\pi \int_0^1 \left( 216 r \sin(\phi) \cos(\theta) r \cos(\phi) + 12 r \sin(\phi) \sin(\theta) + 162 (r \cos(\phi))^2 \right) \\ r^2 \sin(\phi) \ \text{d} r \text{d} \theta \text{d} \phi \end{align}

So it breaks down to solving this integral. Is everything correct so far? I would be grateful to know. Cheers

• Could anyone please confirm my taken approach given the answer by @heropup? – MJimitater Jul 13 '20 at 9:15

Perform the scaling transformation $$(x,y,z) \mapsto (2u, v, 3w)$$ and compute the Jacobian of this transformation. The region of integration becomes a unit sphere in the $$(u,v,w)$$ coordinate system, and the integrand becomes $$2v + 36uw + 27w^2$$. Then a transformation to spherical coordinates is performed as usual.