Question

Sketch the solid described by x 2 + y 2 ≤ z ≤ 1. Use the Divergence Theorem to evaluate the surface integral over the boundary of that solid of the vector field −→F (x, y, z) = y −→ı + z −→ + xz −→k .

Answer 1

Looks like

$\vec F(x,y,z)=y\,\vec\imath+z\,\vec\jmath+xz\,\vec k$

The solid described by $x^2+y^2\le z\le 1$ is the interior of the region bounded by the paraboloid $z=x^2+y^2$ and the plane $z=1$. Call this region $R$.

By the divergence theorem,

$\displaystyle\iint_{\partial R}\vec F\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F\,\mathrm dV$

The divergence is

$\mathrm{div}\vec F(x,y,z)=(y)_x+(z)_y+(xz)_z=x$

so that the surface integral reduces to

$\displaystyle\iiint_Rx\,\mathrm dV$

or in cylindrical coordinates,

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1r^2\cos\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{0}$