Section 15.6: Triple Integration

SOLUTION We could use any of the six possible orders of integration. If we choose to integrate with respect to \(x\), then \(y\), and then \(z\), we obtain \[ \iiint_B xyz^2 dV = \int_0^3 \int_{-1}^2 \int_0^1 xyz^2 dx dy dz = \int_0^3 \int_{-1}^2 \left[ \frac{x^2yz^2}{2} \right]_{x=0}^{x=1} dy dz \] \[ = \int_0^3 \int_{-1}^2 \frac{yz^2}{2} dy dz = \int_0^3 \left[ \frac{y^2z^2}{4} \right]_{y=-1}^{y=2} dz = \int_0^3 \frac{3z^2}{4} dz = \left[ \frac{z^3}{4} \right]_0^3 = \frac{27}{4} \]