Chapter 12: Q50E (page 730)

Find the average value of the function\({\rm{f(x,y,z) = }}{{\rm{x}}^{\rm{2}}}{\rm{z + }}{{\rm{y}}^{\rm{2}}}{\rm{z}}\)over the region enclosed by the parabolic\({\rm{z = 1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}\)and the plane z=0.

Short Answer

Expert verified

The average value of the function \({\rm{f(x,y,z) = }}{{\rm{x}}^{\rm{2}}}{\rm{z + }}{{\rm{y}}^{\rm{2}}}{\rm{z}}\)is \(\frac{{\rm{1}}}{{{\rm{12}}}}\).

Step by step solution

Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

Find polar coordinates

The intersection of the parabolic and the\(xy\)plane is obtained by setting z=0. This is clearly a circle, as\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 1}}\). We may easily express y to set the bounds of the second integral using the circle equation. The volume is calculated by taking the integral of 1 over the given region:

\({\rm{V = }}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{\rm{ - }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} } {\int_{\rm{0}}^{{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} {\rm{d}} } } {\rm{zdydx = }}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{\rm{ - }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} } {\rm{1}} } {\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}{\rm{dydx}}\)

To solve the double integral, switch to polar coordinates:

\(\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\rm{r}} } {\rm{ - }}{{\rm{r}}^{\rm{3}}}{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{2}}}\)