Evaluate the triple integral of yz over the region 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2, y² ≤ z ≤ 4.

Step-by-step solution

1. 1

   Check that the limits are consistent. For $0 \le y \le 2$ we have $y^2 \le 4$, so the inner range $y^2 \le z \le 4$ is non-empty throughout, collapsing to a point only at $y = 2$. The region is the solid between the parabolic surface $z = y^2$ and the plane $z = 4$, extruded along $x$.

2. 2

   Integrate in z, now with a variable lower limit.

   $\int_{y^2}^{4} yz\,dz = y\left[\frac{z^2}{2}\right]_{y^2}^{4} = y\left(\frac{16}{2} - \frac{y^4}{2}\right) = 8y - \frac{y^5}{2}$

   Squaring the lower limit gives $(y^2)^2 = y^4$; forgetting to square it would leave $y^2$ and change the answer.

3. 3

   Integrate in y from 0 to 2.

   $\int_0^2 \left(8y - \frac{y^5}{2}\right) dy = \left[4y^2 - \frac{y^6}{12}\right]_0^2 = 16 - \frac{64}{12} = 16 - \frac{16}{3} = \frac{32}{3}$

4. 4

   Integrate in x — the integrand no longer depends on x. Since nothing in the reduced expression involves $x$, this step is just multiplication by the length of the $x$-interval:

   $\int_0^{2\pi} \frac{32}{3}\,dx = \frac{32}{3}\cdot 2\pi = \frac{64\pi}{3}$

5. 5

   State and sanity-check the answer.

   $\iiint yz\,dV = \frac{64\pi}{3} \approx 67.02$

   Compare with the companion problem where $z$ runs from $0$ to $y^2$: that gives $\frac{32\pi}{3}$, and the two together must sum to the integral over the full box $0 \le z \le 4$, namely $2\pi \cdot \int_0^2 y\,dy \cdot \int_0^4 z\,dz = 2\pi \cdot 2 \cdot 8 = 32\pi$. Indeed $\frac{64\pi}{3} + \frac{32\pi}{3} = 32\pi$ — a complete consistency check.