Finding the Bounds for Evaluating a Triple Integral

[carl123]

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

 

[Prove It]

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

Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars would be the best to use. So your bounds are $\displaystyle \begin{align*} 5\,y^2 + 5\,z^2 \leq x \leq 5 \implies 5\,r^2 \leq x \leq 5 \end{align*}$, since each cross-section is a full circle $\displaystyle \begin{align*} 0 \leq \theta \leq 2\pi \end{align*}$. Note that when $\displaystyle \begin{align*} x = 0 \implies y^2 + z^2 = 0 \implies r = 0 \end{align*}$ and $\displaystyle \begin{align*} x = 5 \implies y^2 + z^2 = 1 \implies r = 1 \end{align*}$, so your bounds for r are $\displaystyle \begin{align*} 0 \leq r \leq 1 \end{align*}$. So finally we can set up our triple integral as

$\displaystyle \begin{align*} \int{\int{\int_E{5\,x\,\mathrm{d}V}}} = \int_0^{2\,\pi}{ \int_0^1{ \int_{5\,r^2}^5{ 5\,x\,r \,\mathrm{d}x } \,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*}$

Now it's up to you to do the integration :)