1. Triple Integral Setup

Problem 11 from chapter 16, section 9 of Calculus with Analytic Geometry by Purcell and Varberg, reads as follows:

Evaluate $\displaystyle \iiint_{S}z^2 dV$, where $\displaystyle S$ is the region bounded by $\displaystyle x^2 + z = 1$ and $\displaystyle y^2 + z = 1$ and the $\displaystyle xy$-plane.

I envisioned the region as a pyramidal paraboloid centered at the origin with a height of 1 and a 2x2 base. I set up the following integral:

$\displaystyle 4\int_0^1\int_0^\sqrt{1 - z}\int_0^\sqrt{1 - z}z^2\,dx\,dy\,dz$

This gave me a result of $\displaystyle \frac{1}{3}$, which WolframAlpha tells me is correct: https://www.wolframalpha.com/input/?...t(1+-+z)%7D%5D

Along the way I had $\displaystyle 4\int_0^1z^2(\sqrt{1-z})^2dz$, which is how I'd have set it up if I knew only single integrals.

However, the answer key says $\displaystyle 0.8857$. I now wonder whether I interpreted the problem properly.

2. Re: Triple Integral Setup

Let $R=$ the region in the $xy$-plane that is inside the triangle with vertices $(0,0,0),(1,1,0),(1,-1,0)$

Over the region $R$ the surface $z_1=1-x^2$ is below the surface $z_2=1-y^2$

so what is the region bounded by these three surfaces $R, z_1, z_2$ ?

The book interprets this as being the region bounded below by the $xy$-plane and above by $z_2$

Therefore the triple integral should be set up as

$\displaystyle \int _0^1\int _{-x}^x\int _0^{1-y^2}z^2dzdydx=\frac{31}{140}$

using symmetry, multiply by $4$ to get the result

$\displaystyle 4\left(\frac{31}{140}\right)=\frac{31}{35}\approx 0.885714$

3. Re: Triple Integral Setup

Thanks, but... the set traverses one of the bounding surfaces?