How to evaluate a Triple Integral

[squenshl]

How would I evaluate the integral $$\int\int\int_{G}$$ $$\sqrt{4x^2+9y^2}$$ dV,
where G is the elliptic cylinder 4x²+9y² $$\leq$$ 25,
0 $$\leq$$ z $$\leq$$ 6

 

[GPPaille]

You can easily integrate for Z, since there's no Z in the integrand. Next, you need to do two changes of variables. The first is simply x'=2x and y'=3y. You will get a very familiar integral (don't forget the jacobian). Then you go in polar coordinates (Mr. Jacob is here too) to integrate a very very simple equation.

 

[squenshl]

I was given a hint:
Try modified cylindrical coordinates with$$\vartheta$$ retaining its usual meanings, but with the meaning of r changed so that r² = 4x² + 9y². You will need to make definitions such as x = a₁r cos$$\vartheta$$
and y = a₂ sin$$\vartheta$$, for some suitable numbers a₁ and a₂. You will need to calculate the jacobian for this mapping.

 

[GPPaille]

Yes you can do it. It's even simpler than what I said but it's exactly the same thing, except that it's in one single step.

So you have your answer or your stuck there?

 

[squenshl]

Just a little bit stuck. I tried it but it didn't seem to work out.

 

[squenshl]

I don't know how to start it. If someone would show me how to do this it would be really appreciated.

 

[GPPaille]

You need to define (x,y,z) with (r,t,z'):

$$x = \frac{r}{2}cos\theta$$

$$y = \frac{r}{3}sin\theta$$

$$z = z'$$

$$\left| J \right| = \begin{vmatrix} \frac{1}{2}cos\theta & -\frac{r}{2}sin\theta & 0 \\ \frac{1}{3}sin\theta & \frac{r}{3}cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = \frac{r}{6}$$

Noting that using this transformation, r is really equal to $\sqrt{4x^2 + 9y^2}$, you finally have this integral:

$$\int_{r=0}^5 { \int_{\theta=0}^{2 \pi} { \int_{z'=0}^6 { r \frac{r}{6}dz' d\theta dr } } }$$

 

[squenshl]

Of course. It's so obvious. Thanks heaps.