Answer to Question #313342 in Calculus for Jhonny

Question

Evaluate ∫∫∫E 6z^2dV where E is the region below 4x + 2y + 2z = 10 in the first octant.

Answer 1

ANSWER $\int\int\int_{E} \ 6z^2dV=$ $\frac{625}{4} =156.25\\$

EXPLANATION

The projection of the set $E=\left \{ \left ( x,y,z \right ):0\leq x\leq\frac{5} {2},\, 0\leq y\leq5-2x,\, 0\leq z\leq 5-2x-y\right \}$

onto the x0y plane is the region

$D=\left \{ \left ( x,y \right ):0\leq x\leq\frac{5} {2},\, 0\leq y\leq5-2x \right \} .$

$\int\int\int_{E} \ 6z^2dV=\int_{0}^{\frac{5}{2}} \int_{0}^{5-2x}\int_{0}^{5-2x-y}6z^2dzdydx=6\int_{0}^{\frac{5}{2}} \int_{0}^{5-2x}\left [ \frac{z^3}{3} \right ]_{0}^{5-2x-y}dydx= 2\int_{0}^{\frac{5}{2}} \int_{0}^{5-2x}\left (5-2x-y \right )^3dydx=2\int_{0}^{\frac{5}{2}}\left [ \frac{\left ( 5-2x-y \right )^{4}}{-4} \right ]_{y=0}^{y=5-2x} dx= \\=\frac{1 }{2}\int_{0}^{\frac{5}{2}}(5-2x)^{4}dx= \frac{1}{2}\left [ \frac{(5-2x)^{5}}{-10} \right ]_{0 }^{\frac{5}{2}}=\frac{5^{5}}{20}=\frac{625}{4} =156.25\\$