Question

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = −xi − yj + z3k, S is the part of the cone z = x2 + y2 between the planes z = 1 and z = 2 with downward orientation.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The equation of the cone should be $z=\sqrt{x^2+y^2}$. Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$. Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k$

Then the integral of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_1^2(-u^2-u^4)\,\mathrm du\,\mathrm dv$

$=\displaystyle-2\pi\int_1^2(u^2+u^4)\,\mathrm du\,\mathrm dv=\boxed{-\frac{256\pi}{15}}$