Section 15.6: Triple Integration

SOLUTION The tetrahedron \(T\) and its projection \(D\) onto the \(xy\)-plane are shown in Figures 14 and 15. The lower boundary of \(T\) is the plane \(z = 0\) and the upper boundary is the plane \(x + 2y + z = 2\), that is, \(z = 2 - x - 2y\). Therefore we have \[ V(T) = \iiint_T dV = \int_0^1 \int_{x/2}^{1-x/2} \int_0^{2-x-2y} dz dy dx \] \[ = \int_0^1 \int_{x/2}^{1-x/2} (2 - x - 2y) dy dx = \frac{1}{3} \] by the same calculation as in Example 15.2.4.