Question

Evaluate the triple integral. T 6x2 dV, where T is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1)

Answer 1

`T` is the set

`T = \left\{(x,y,z) \mid 0 \le x \le 1 \text{ and } 0 \le y \le 1 - x \text{ and } 0 \le z \le 1 - x - y\right\}`

since the plane
`x+y+z=1` passes through the three, non-origin vertices.

Then the integral of
`6x^2` over
`T` is

`\displaystyle \iiint_T 6x^2 \, dV = \int_0^1 \int_0^(1-x) \int_0^(1-x-y) 6x^2 \, dz \, dy \, dx \\\\ ~~~~~~~~ = 6 \int_0^1 \int_0^(1-x) x^2 (1-x-y) \, dy \, dx \\\\ ~~~~~~~~ = 6 \int_0^1 \int_0^(1-x) (x^2 - x^3 - x^2y) \, dy \, dx \\\\ ~~~~~~~~ = 6 \int_0^1 \left((x^2 - x^3) (1-x) - \frac{x^2}2 (1-x)^2\right) \, dx \\\\ ~~~~~~~~ = 6 \int_0^1 \left(\frac{x^4}2 - x^3 + \frac{x^2}2\right) \, dx \\\\ ~~~~~~~~ = 6 \left(\frac1{10} - \frac14 + \frac16\right) = \boxed{\frac1{10}}`