Question

Let \(\mathcal{C}\) be the set of points interior to or on the boundary of a cube with edge of length 1. Moreover, say that the cube is in the first octant with one vertex at the point \((0,0,0)\) and an opposite vertex at the point \((1,1,1)\). Let \(Q(C)=\) \(\iiint_{C} d x d y d z\) (a) If \(C \subset \mathcal{C}\) is the set \(\\{(x, y, z): 0

Short answer

The value of \(Q(C)\) for region (a) is \(1/2\) and (b) is \(0\).

Step-by-step solution

Defining the Integral Limits for Part (a)

For the region \(C = {(x, y, z): 0<x<y<z<1}\), the integral iterates over \(z\) first, then \(y\), then \(x\). We find the limits as \(x\) varies from 0 to 1, \(y\) varies from \(x\) to 1 and \(z\) varies from \(y\) to 1.

Evaluate Integral for Part (a)

Having established the limits, the triple integral calculation for \(Q(C)\) can be written as: \[ Q(C) = \int_{0}^{1} \int_{x}^{1} \int_{y}^{1} dz dy dx \] Because there is no function before \(dz \, dy \, dx\), this is merely calculating the volume of the region, i.e., it's equivalent to \(1 \times \) volume. The integral of \(dz\) is \(z\), evaluated from \(y\) to 1, which yields \(1 - y\). Next we integrate \(1 - y\) with respect to \(y\), from \(x\) to 1, which yields \(y - (y^2)/2\), yielding \(1 - x - 1/2 \, (1 - x^2)\). Finally, we integrate with respect to \(x\) from 0 to 1, which yields \(1/2\).

Defining the Integral Limits for Part (b)

For the region \(C = {(x, y, z): 0<x=y=z<1}\), our limits now collapse to the line \(x = y = z\). Thus, \(x\) varies from 0 to 1.

Evaluate Integral for Part (b)

Having established the limits, the triple integral calculation for \(Q(C)\) can be written as: \[ Q(C) = \int_{0}^{1} dx \] This integral is again equivalent to volume, but in this case it evaluates to zero as it represents a line within the cube.