Question

Evaluate the triple integral ∭Tx2dV, where T is the solid tetrahedron with vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3).

Answer 1

Answer:

the integral I=81

Step-by-step explanation:

for the integral I

$I=\int\limits^{}_{}\int\limits^{}_{}\int\limits^{}_{T} {x^{2} } \, dxdydz$

where T is the solid tetrahedron , then

$I=\int\limits^{3}_{0}\int\limits^{3}_{0}\int\limits^{3}_{0} {x^{2} } \, dxdydz = \int\limits^{3}_{0}dz\int\limits^{3}_{0}dy\int\limits^{3}_{0} {x^{2} } \, dx = (3-0)*(3-0)*1/3*(3^{3}-0^{3}) = 3^{4} = 81$

the integral is equal to 81