# Evaluate the triple integral

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1. Oct 21, 2015

### carl123

Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid
x = 5y2+ 5z2 and the plane x = 5.

My work so far:
Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y2+5z2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Also, when x=0, y2 + z2=0 -----> r=0 and x=5 -----> y2+z2=1 -----> r=1, so my bounds for r are 0 ≤r ≤ 1.

My triple integral is then:
∫∫∫E 5x dV = ∫(from 0 to 2π) ∫(from 0 to 1) ∫(from 5r2 to 5) 5xr dx dr dθ

I keep getting the wrong answer after evaluating it

2. Oct 21, 2015

### Ray Vickson

What answer do you get when you evaluate it? What do you think is the correct answer?

3. Oct 22, 2015

### geoffrey159

Your domain of integration is $E=\{(x,y,z) \in\mathbb{R}^3, \ 0\le x \le 5, 0\le y^2 + z^2 \le x/5\}$

If you sketch what it looks like, you'll see that a natural change of variable is $(x,y,z) = \phi(x',\rho,\theta) = (x', \rho \cos\theta, \rho \sin\theta)$, with $\rho > 0$ and $\theta \in [0,2\pi[$ to make it bijective except for the points on the $x$ axis.

Then, your reciprocal domain is $\phi^{-1}(E) = \{ 0\le x' \le 5,\ 0<\rho\le \sqrt{x'/5}, \ 0\le \theta < 2\pi \}$

4. Oct 22, 2015

### Staff: Mentor

Please don't delete the three parts of the homework template. Its use is required.