Question

Evaluate the triple integral


B
xyz2 dV,
where B is the rectangular box
B =

(x, y, z) | 0 ≤ x ≤ 1, −1 ≤ y ≤ 7, 0 ≤ z ≤ 4

,
integrating first with respect to y, then z, and then x.

Answer 1

Each face of the box B is described completely by a single variable, so the value of the integral is independent of the order of integration.

The integral in question would be

$\displaystyle \iiint_B xyz^2 \, dV = \int_0^1 \int_0^4 \int_{-1}^7 xyz^2 \, dy \, dz \, dx$

but we can factorize the integral as

$\displaystyle \iiint_B xyz^2 \, dV = \left(\int_0^1 x \, dx\right) \left(\int_0^4 z^2 \, dz\right) \left(\int_{-1}^7 y \, dy\right)$

We can do this regardless of the order of integration we choose. Then

$\displaystyle \int_0^1 x \, dx = \frac12(1^2-0^2) = \frac12$

$\displaystyle \int_0^4 z^2 \, dz = \frac13 (4^3 - 0^3) = \frac{64}3$

$\displaystyle \int_{-1}^7 y \, dy = \frac12 (7^2-(-1)^2) = 24$

and the triple integral has a value of

$\displaystyle \iiint_B xyz^2 \, dV = \frac12 \cdot \frac{64}3 \cdot 24 = \boxed{256}$

But, if you insist on carrying out just one integral at a time, we have

$\displaystyle \iiint_B xyz^2 \, dV = \int_0^1 \int_0^4 \int_{-1}^7 xyz^2 \, dy \, dz \, dx$

$\displaystyle \iiint_B xyz^2 \, dV = \int_0^1 \int_0^4 \frac12 xz^2 (7^2-(-1)^2) \, dz \, dx = 24 \int_0^1 \int_0^4 xz^2 \, dz \, dx$

$\displaystyle \iiint_B xyz^2 \, dV = 24 \int_0^1 \frac13 x(4^3 - 0^3) \, dx = 512 \int_0^1 x \, dx$

$\displaystyle \iiint_B xyz^2 \, dV = 512 \cdot \frac12 (1^2 - 0^2) = \boxed{256}$