valuate the triple integral ∭xydV where the region is the solid tetrahedon with vertices (0,0,0), (6,0,0), (0,10,0), (0,0,9)

Answer 1

One way to do this is to think of the solid as a stack of right triangular slices perpendicular to the z axis.

For a right triangle of height H and base W, the integration of x y dx dy over the triangle can be shown to be

(1/24) H^2 W^2.

As z is increased from 0 to 9, the relevant H and W scale as

H ==> H ( 9 - z)/9 and W ==> W (9 - z)/9

Thus the integration over z has the integrand (1/24) H^2 W^2 [ (9 - z)/9 ]^4 with upper limit 9 and lower limit 0. (Here H = 9 W = 6)

This integral is easy to evaluate. The result is (1/120) 9 H^2 W^2

Plugging in W = 6 and H = 10 gives 270