How do I evaluate $\iiint_{R} (xy+z^2) \ dV\,$?

Question

The question states:

Sketch the region $$R = \left \{ (x,y,z) : 0 \leq z \leq 1- |x| -|y| \right \}$$

and evaluate the following integral $$\iiint_{R} (xy+z^2) \ dV$$

So my thoughts are that the sketch is straightforward just a pyramid with a square base. But then I get stuck with the limits of the integral. All I know is that we can take the positive segments and double and add on to find R since it's symmetrical but if someone could explain the limits involved.

The answer has been given to be: $$\frac{1}{15}$$

Thank you!

Answer 1

The integral $\int_R xy\>dV=0$ because of symmetry.

The base square of $R$ has side length $\sqrt{2}$. When $R$ is intersected with a horizontal plane at height $z\in[0,1]$ we therefore obtain a square of area $\bigl((1-z)\sqrt{2}\bigr)^2$. It follows that $$\int_R z^2\>dV=\int_0^1 z^2\>2(1-z)^2 \>dz={1\over15}\ .$$

Answer 2

You are right about the shape of $R$. Since$$|x|+|y|\leqslant1\iff|y|\leqslant1-|x|\iff|x|-1\leqslant y\leqslant1-|x|,$$your integral is$$\int_{-1}^1\int_{|x|-1}^{1-|x|}\int_0^{1-|x|-|y|}(xy+z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$$