Evaluate the triple integral of 4x^2*z^3*y with x from 0 to 1, y from 1 to 2 and z from 1 to 2.

Step-by-step solution

1. 1

   Read the order of integration off the differentials. The differentials appear as $dx\,dy\,dz$, so the innermost variable is $x$ (limits $0$ to $1$), then $y$ (limits $1$ to $2$), then $z$ (limits $1$ to $2$). All six limits are constants, so this is an integral over a rectangular box and no boundary curves are involved.

2. 2

   Integrate in x. Treating $z^{3}$ and $y$ as constants, $\int_{0}^{1}4x^{2}z^{3}y\,dx=4z^{3}y\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{4}{3}z^{3}y .$

3. 3

   Integrate in y. Now $\frac43 z^{3}$ is constant with respect to $y$: $\int_{1}^{2}\frac{4}{3}z^{3}y\,dy=\frac{4}{3}z^{3}\left[\frac{y^{2}}{2}\right]_{1}^{2}=\frac{4}{3}z^{3}\cdot\frac{3}{2}=2z^{3}.$

4. 4

   Integrate in z. $\int_{1}^{2}2z^{3}\,dz=\left[\frac{z^{4}}{2}\right]_{1}^{2}=\frac{16}{2}-\frac{1}{2}=\frac{15}{2}.$

5. 5

   Cross-check by separating the factors. Because the integrand factors as $4\cdot x^{2}\cdot y\cdot z^{3}$ and the box has constant limits, the integral is the product of one-variable integrals: $4\cdot\frac13\cdot\frac32\cdot\frac{15}{4}=\frac{15}{2}$, matching the iterated computation.