# Evaluate the integrals by making a change of coordinates. I= \int_{0}^{1}...

## Question:

Evaluate the integrals by making a change of coordinates.

{eq}I= \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{2-x^2-y^2} z(x^2+y^2+z^2) dzdydx {/eq}

## Cylindrical Coordinates:

One of the most valuable skills we can develop to help us when working with integrals when to use what coordinate system. In the integral above, there are lots of sums of two squares. This is a good indication that we should use cylindrical coordinates. Recall

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

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

{eq}z = z {/eq}

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

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

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

## Answer and Explanation:

We will be using cylindrical coordinates. Note that the bounds on {eq}x {/eq} and {eq}y {/eq} describe that part of the unit circle that lies in quadrant 1, so {eq}r \leq 1 {/eq} and {eq}\theta \in [0, \frac\pi4] {/eq}. The bounds on {eq}z {/eq} become {eq}r \leq z \leq 2-r^2 {/eq} and so we find

{eq}\begin{align*} \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{2-x^2-y^2} z(x^2+y^2+z^2) \ dz\ dy\ dx &= \int_0^{\pi/2} \int_0^1 \int_r^{2-r^2} z(r^2+z^2)\ r\ dz\ dr\ d\theta \\ &= \int_0^{\pi/2}d \theta \int_0^1 \int_r^{2-r^2} r^3z + rz^3\ dz\ dr \\ &= \left( \frac\pi2 \right) \int_0^1 \left [ \frac12 r^3 z^2 + \frac14 r z^4 \right ]_r^{2-r^2}\ dr \\ &= \frac\pi2 \int_0^1 \frac12 r^3 (2-r^2)^2 - \frac12 r^3(r)^2 + \frac14r(2-r^2)^4 - \frac14 r(r)^4 \ dr \\ &= \frac\pi2 \int_0^1 \frac12 r^3 (2-r^2)^2 + \frac14r(2-r^2)^4 - \frac34 r^5 \ dr \\ &= \frac\pi2 \left[ \frac12 \left( \frac18 r^8 - \frac23r^6 + r^4 \right) - \frac1{4\cdot 10} \left( 2-r^2 \right)^5 - \frac18r^6 \right]_0^1 \\ &= \frac\pi2 \left[ \frac12 \left( \frac18 - \frac23+ 1 \right) - \frac1{40} \left( 2-1 \right)^5 + \frac1{40} \left( 2 \right)^5 - \frac18 \right] \\ &= \frac{211\pi}{480} \\ &\approx 1.381 \end{align*} {/eq}

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from GRE Math: Study Guide & Test Prep

Chapter 13 / Lesson 10