Evaluate the triple integral of xz over the region −1 ≤ x ≤ 1, x² ≤ y ≤ 1, 0 ≤ z ≤ y.

Step-by-step solution

1. 1

   Read the nesting order off the limits. The bounds are already stacked correctly for $dz\,dy\,dx$: the innermost limits ($0$ to $y$) may involve $y$ and $x$, the middle ones ($x^2$ to $1$) may involve $x$, and the outer ones must be constants. No reordering is needed.

2. 2

   Integrate in z. Here $x$ is a constant:

   $\int_0^y xz\,dz = x\left[\frac{z^2}{2}\right]_0^y = \frac{xy^2}{2}$

3. 3

   Integrate in y from x² to 1. Now $x/2$ is the constant:

   $\int_{x^2}^{1} \frac{xy^2}{2}\,dy = \frac{x}{2}\left[\frac{y^3}{3}\right]_{x^2}^{1} = \frac{x}{6}\left(1 - x^6\right)$

   Note $(x^2)^3 = x^6$, not $x^5$ — exponents multiply when a power is raised to a power.

4. 4

   Look at the parity of the remaining integrand before integrating.

   $\frac{x(1 - x^6)}{6} = \frac{x - x^7}{6}$

   Both $x$ and $x^7$ are odd functions, so the whole integrand is odd: $f(-x) = -f(x)$.

5. 5

   Use the symmetry of the interval. The outer interval $[-1, 1]$ is symmetric about $0$, and the integral of any odd function over such an interval is zero:

   $\frac{1}{6}\int_{-1}^{1}(x - x^7)\,dx = 0$

   The geometric reason is visible in the setup: the solid region is symmetric under $x \mapsto -x$ (the bounds $x^2 \le y \le 1$ and $0 \le z \le y$ are unchanged), while the integrand $xz$ flips sign, so contributions cancel in pairs.

6. 6

   Verify. Direct evaluation term by term gives $\int_{-1}^{1} x\,dx = 0$ and $\int_{-1}^{1} x^7\,dx = 0$. Numerical triple quadrature returns $-5 \times 10^{-20}$, i.e. zero to machine precision. Spotting the symmetry at the outer stage saves the whole computation — but only after you check that the region really is symmetric.