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

Step-by-step solution

1. 1

   Describe the solid. The inequality $x^2 + y^2 \le z$ says the point lies inside the paraboloid $z = x^2 + y^2$, which opens upward from the origin. Capping it at $z = 3$ gives a bowl-shaped solid whose cross-section at height $z$ is the disk $x^2 + y^2 \le z$ of radius $\sqrt{z}$.

2. 2

   Test the region for symmetry in x. Replacing $x$ by $-x$ leaves both defining conditions untouched, since only $x^2$ appears:

   $(-x)^2 + y^2 = x^2 + y^2 \le z$

   So $(x, y, z) \in E \iff (-x, y, z) \in E$: the solid is symmetric about the plane $x = 0$. This check is essential — the shortcut fails on a region that is not symmetric.

3. 3

   Test the integrand for parity in x. The integrand is $f(x,y,z) = x$, and

   $f(-x, y, z) = -x = -f(x,y,z)$

   so it is odd in $x$. An odd integrand over a region symmetric in that variable integrates to zero, because every point pairs with a mirror point carrying the opposite value.

4. 4

   Conclude without computing.

   $\iiint_E x\,dV = 0$

   The positive contributions from the half $x > 0$ exactly cancel the negative ones from $x < 0$.

5. 5

   Confirm by setting up the cylindrical integral anyway. With $x = r\cos\theta$ and $dV = r\,dz\,dr\,d\theta$:

   $\int_0^{2\pi}\!\!\int_0^{\sqrt{3}}\!\!\int_{r^2}^{3} r^2\cos\theta\,dz\,dr\,d\theta$

   The $\theta$-integral factors out as $\int_0^{2\pi}\cos\theta\,d\theta = 0$, so the whole product is zero regardless of the radial and vertical parts. Same answer, now confirmed algebraically. (By the same argument $\iiint_E y\,dV = 0$ and $\iiint_E xyz\,dV = 0$, while $\iiint_E z\,dV$ is genuinely nonzero.)