Evaluate the triple integral of xyz over the solid region where 0 ≤ z ≤ 3 and x² + y² ≤ z.

Step-by-step solution

1. 1

   Describe the solid. The condition $x^{2}+y^{2}\le z$ places the point inside the paraboloid $z=x^{2}+y^{2}$, and $z\le 3$ caps it with a plane. At height $z$ the cross-section is a disc of radius $\sqrt z$, so the solid is a bowl of depth $3$ and top radius $\sqrt3$.

2. 2

   Switch to cylindrical coordinates. With $x=r\cos\theta$, $y=r\sin\theta$ and $dV=r\,dz\,dr\,d\theta$,

   $xyz=r^{2}\cos\theta\sin\theta\cdot z,$

   so

   $\iiint_E xyz\,dV=\int_{0}^{2\pi}\!\int_{0}^{\sqrt3}\!\int_{r^{2}}^{3} r^{3}z\cos\theta\sin\theta\,dz\,dr\,d\theta.$

3. 3

   Notice the integral separates. Nothing in the limits depends on $\theta$, and the $\theta$-dependence of the integrand is the single factor $\cos\theta\sin\theta$. The whole triple integral therefore factors as

   $\left(\int_{0}^{2\pi}\cos\theta\sin\theta\,d\theta\right)\left(\int_{0}^{\sqrt3}\!\int_{r^{2}}^{3}r^{3}z\,dz\,dr\right).$

   Checking for this before computing anything saves the entire radial calculation.

4. 4

   Evaluate the angular factor. Using $\cos\theta\sin\theta=\tfrac12\sin 2\theta$,

   $\int_{0}^{2\pi}\frac{\sin 2\theta}{2}\,d\theta=\left[-\frac{\cos 2\theta}{4}\right]_{0}^{2\pi}=-\frac14+\frac14=0.$

5. 5

   Conclude, and see why geometrically. Since the first factor is $0$ and the second is finite, the whole integral is

   $\iiint_E xyz\,dV=0.$

   The reason is symmetry: the solid is unchanged by $x\mapsto-x$, while $xyz$ changes sign, so contributions cancel in pairs. Numerical triple quadrature over the region returns $-9.6\times10^{-16}$, i.e. zero to machine precision.