Answer to Question #315260 in Calculus for aho

Question

The temperature at a point is given by T = xyz. Find the average temperature in the cube

with opposite corners at (0,0,0) and (2,2,2).

Answer 1

$T=xyz\\ (0,0,0), (2,2,2)\\ \frac{1}{2\cdot2\cdot2} \int_0^2\int_0^2\int_0^2xyzdxdydz=\frac{1}{8}\int_0^2\int_0^2xy\frac{z^2}{2}|_0^2dxdy=\\ =\frac{1}{8}\int_0^2\int_0^22xydxdy=\frac{1}{4}\int_0^2x\frac{y^2}{2}|_0^2dx=\frac{1}{4}\int_0^22xdx=\\ =\frac{1}{2}\frac{x^2}{2}|_0^2=\frac{1}{2}\cdot2=1$