# Evaluate the triple integral triple integral_E (x + 1) dV, where E is the bounded by the...

## Question:

Evaluate the triple integral

{eq}\displaystyle \iiint_E (x + 1)\ dV {/eq}, where E is the bounded by the paraboloid {eq}\displaystyle x = y^2 + z^2 {/eq} and the plane {eq}x = \pi^2 {/eq}.

## Cylindrical Coordinates:

Given the situation above, we immediately want to use cylindrical coordinates. But we will need to change things a bit from the usual way we see them sice we need the axis of rotation to be {eq}x {/eq} instead of {eq}z {/eq}. We will be using the following:

{eq}x=x {/eq}

{eq}y = r \cos \theta {/eq}

{eq}z = r \sin \theta {/eq}

{eq}r^2 = y^2+z^2 {/eq}

{eq}\theta = \tan^{-1} \frac{z}{y} {/eq}

{eq}dV = r \ dx \ dr \ d\theta {/eq}

## Answer and Explanation: 1

With our adjusted cylindrical coordinates, the given surfaces bound {eq}x {/eq} as {eq}r^2 \leq x \leq \pi^2 {/eq}. Then, we bound {eq}r {/eq} at the intersection of these surfaces.

{eq}\begin{align*} r^2 &= \pi^2 \\ r &= \pi \end{align*} {/eq}

We need to cover the entire paraboloid so {eq}\theta \in [0,2\pi] {/eq} and we get

{eq}\begin{align*} \iiint_E (x+1)\ dV &= \int_0^{2\pi} \int_0^{\pi} \int_{r^2}^{\pi^2} (x+1)\ r\ dx\ dr\ d\theta \\ &= \int_0^{2\pi} d \theta \int_0^{\pi} r \left [ \frac12x^2+x \right ]_{r^2}^{\pi^2}\ dr \\ &= (2\pi) \int_0^{\pi} r \left( \frac12(\pi^4-r^4) + \pi^2-r^2 \right)\ dr \\ &= 2\pi \int_0^{\pi} \left( \frac12\pi^4 + \pi^2 \right) r - \frac12 r^5 - r^3\ dr \\ &= 2\pi \left [ \frac12\left( \frac12\pi^4 + \pi^2 \right)r^2 - \frac1{12}r^6 - \frac14r^4 \right ]_0^\pi \\ &= \pi \left [ \left( \frac12\pi^4 + \pi^2 \right)\pi^2 - \frac1{6}(\pi)^6 - \frac12(\pi)^4 \right ] \\ &= \frac{\pi^5}{6}(3+2\pi^2) \\ &\approx 1159.77 \end{align*} {/eq}