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

Step-by-step solution

1. 1

   See why the order of integration is forced here. The inner $z$-limit is $y^2$, so $z$ must be integrated before $y$ — the middle variable cannot be eliminated while the inner limit still depends on it. The region is the solid under the parabolic sheet $z = y^2$, extruded along $x$.

2. 2

   Integrate in z. Holding $y$ constant:

   $\int_0^{y^2} yz\,dz = y\left[\frac{z^2}{2}\right]_0^{y^2} = y \cdot \frac{y^4}{2} = \frac{y^5}{2}$

   The integrand $y$ and the limit $y^2$ combine, pushing the power from $1$ up to $5$.

3. 3

   Integrate in y from 0 to 2.

   $\int_0^2 \frac{y^5}{2}\,dy = \frac{1}{2}\left[\frac{y^6}{6}\right]_0^2 = \frac{64}{12} = \frac{16}{3}$

4. 4

   Integrate in x. Nothing left depends on $x$, so this multiplies by the interval length $2\pi$:

   $\int_0^{2\pi}\frac{16}{3}\,dx = \frac{16}{3}\cdot 2\pi = \frac{32\pi}{3}$

5. 5

   Check against the complementary region. Over the full box $0 \le z \le 4$ the integral is $2\pi \cdot \int_0^2 y\,dy \cdot \int_0^4 z\,dz = 2\pi \cdot 2 \cdot 8 = 32\pi$. The part above the sheet, $y^2 \le z \le 4$, works out to $\frac{64\pi}{3}$, and

   $\frac{32\pi}{3} + \frac{64\pi}{3} = 32\pi \ \checkmark$

   So $\frac{32\pi}{3} \approx 33.51$ is confirmed.