Question

Evaluate the triple integral.

Answer 1

The definition of the set E gives you a natural choice for the limits in the integral:

$\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x \int_{x-y}^{x+y} y \, dz \, dy \, dx$

Computing the integral, we get

$\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x y ((x+y)-(x-y)) \, dy \, dx = 2 \int_0^6 \int_0^x y^2 \, dy \, dx$

$\displaystyle \iiint_E y \, dV = 2 \int_0^6 \frac13 (x^3 - 0^3) \, dx = \frac23 \int_0^6 x^3 \, dx$

$\displaystyle \iiint_E y \, dV = \frac23 \cdot \frac14 (6^4 - 0^4) = \boxed{216}$