Question

Evaluate the triple integral ∭E z dV where E is the solid bounded by the cylinder y^2+z^2=576 and the planes x=0,y=4x and z=0 in the first octant.

Answer 1

$\iiint_E zdV,$

$E=\{y^2+z^2 = 256, y=4x,x=0,z=0, x>0,y>0,z>0\}$

$z \in[0;\sqrt{256-y^2}]\\ x\in[0;\frac{y}{4}]\\ y\in[0;16]$

$\int_0^{16}dy\int_0^{\frac{y}{4}}dx\int_0^{\sqrt{256-y^2}}zdz = \\ =\int_0^{16}dy\int_0^{\frac{y}{4}}dx(\frac{z^2}{2}|_0^{\sqrt{256-y^2}}) = \\= \int_0^{16}dx\int_0^{\frac{y}{4}}(\frac{256-y^2}{2})dx = \int_0^{16}(128-\frac{y^2}{2})\frac{y}{4}dy=\\= \int_0^{16}(32y-\frac{y^3}{8})dy = 16y^2 - \frac{y^4}{32}|_0^{16} = 4096-2048=\\= 2048$