Evaluate the triple integral of yz over the region 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2, 0 ≤ z ≤ x².

Step-by-step solution

1. 1

   Notice which variable controls the inner limit. The $z$-range runs from $0$ to $x^2$ — it depends on $x$, the outermost variable, not on $y$. That is legal, and it means the $y$-integration can be done independently of $z$ once the inner integral is evaluated.

2. 2

   Integrate in z with y treated as a constant.

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

   Squaring the upper limit gives $x^4$, not $x^2$ — this is the step where the final power of $\pi$ is decided.

3. 3

   Integrate in y from 0 to 2. Now $x^4/2$ is constant:

   $\int_0^2 \frac{x^4 y}{2}\,dy = \frac{x^4}{2}\left[\frac{y^2}{2}\right]_0^{2} = \frac{x^4}{2}\cdot 2 = x^4$

4. 4

   Integrate in x from 0 to 2π.

   $\int_0^{2\pi} x^4\,dx = \left[\frac{x^5}{5}\right]_0^{2\pi} = \frac{(2\pi)^5}{5}$

5. 5

   Simplify the power. Since $(2\pi)^5 = 2^5\pi^5 = 32\pi^5$:

   $\iiint yz\,dV = \frac{32\pi^5}{5} \approx 1958.53$

   The $2^5 = 32$ must be expanded — writing $2\pi^5/5$ instead would be off by a factor of $32$. Note that $\pi$ plays no trigonometric role here; it is simply the numeric upper limit $2\pi \approx 6.2832$.