Question

evaluate the triple integral. t 4x2 dv, where t is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1)

Answer 1

The region T corresponds to the area under the plane x + y + z = 1 in the first quadrant. So we have, for instance,

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

which depends on the order of integration we choose; in this case, I use dz dy dx.

Then the integral is

$\displaystyle \iiint_T 4x^2 \, dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 4x^2 \, dz \, dy \, dx$

$\displaystyle \iiint_T 4x^2 \, dV = \int_0^1 \int_0^{1-x} 4x^2z \bigg|_0^{1-x-y} \, dy \, dx$

$\displaystyle \iiint_T 4x^2 \, dV= \int_0^1 \int_0^{1-x} 4x^2(1-x-y) \, dy \, dx$

$\displaystyle \iiint_T 4x^2 \, dV= \int_0^1 \int_0^{1-x} (4x^2-4x^3-4x^2y) \, dy \, dx$

$\displaystyle \iiint_T 4x^2 \, dV = \int_0^1 (4x^2y-4x^3y-2x^2y^2) \bigg|_0^{1-x} \, dx$

$\displaystyle \iiint_T 4x^2 \, dV = \int_0^1 (4x^2(1-x)-4x^3(1-x)-2x^2(1-x)^2) \, dx$

$\displaystyle \iiint_T 4x^2 \, dV = \int_0^1 (2x^4 - 4x^3 + 2x^2) \, dx$

$\displaystyle \iiint_T 4x^2 \, dV = \left(\frac25 x^5 - x^4 + \frac23 x^3\right) \bigg|_0^1$

$\displaystyle \iiint_T 4x^2 \, dV = \frac25 - 1 + \frac23 = \boxed{\frac1{15}}$