\overline{T} = \frac{1}{2 ^{3}} \int_{0}^{2} \int_{0}^{2} \int_{0}^{2} x yz d x d y d z = \frac{1}{8} (\int_{0}^{2} x d x)^{3} .

Each one-dimensional integral is two, giving

\overline{T} = 1.

The number is in the temperature units implicit in the given field law and coordinate units.

ii

Solution

For a general positive side length

s

, the cube has volume

s^{3}

, while separability of the integral gives

\int_{0}^{s} \int_{0}^{s} \int_{0}^{s} x yz d x d y d z = (\frac{s ^{2}}{2})^{3} = \frac{s ^{6}}{8} .

Thus the requested general dependence of the mean temperature is

\overline{T} (s) = \frac{s ^{3}}{8} .

For the cube with side length three this gives

\overline{T} (3) = \frac{27}{8} = 3.375.

Equivalently each coordinate has volume-average value

s /2

, and the rectangular integration domain makes the mean of

x yz

the product

(s /2)^{3}

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