Answer
$\frac{640}{3}$
Work Step by Step
Find the domain of $z=16-x^2$ such that the surface lies in the first octant.
$z\geq 0$
$16-x^2\geq 0$
$x^2\leq 16$
$-4\leq x\leq 4$ (In the first octant $x\geq 0$)
$0\leq x\leq 4$
Then, the volume of the solid in the first octant bounded by the cylinder $z=16-x^2$ and the plane $y=5$ is represented by the multiple integral $V:\int_0^5\int_0^416-x^2dxdy$.
Find the volume:
$V=\int_0^5\int_0^416-x^2dxdy=\int_0^516x-\frac{x^3}{3}]_0^4dy=\int_0^516\cdot 4-\frac{4^3}{3}-0dy=\int_0^5\frac{128}{3}dy=[\frac{128y}{3}]_0^5=\frac{128\cdot 5}{3}-0=\frac{640}{3}$
