Question

Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = ey tan(z)i + y 6 − x2 j + x sin(y)k, S is the surface of the solid that lies above the xy-plane and below the surface z = 2 − x4 − y4, −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

Answer 1

$\vec F(x,y,z)=e^y\tan z\,\vec\imath+(y^6-x^2)\,\vec\jmath+x\sin y\,\vec k$

has divergence

$\nabla\cdot\vec F(x,y,z)=6y^5$

Then by the divergence theorem, the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_{-1}^1\int_{-1}^1\int_0^{2-x^4-y^4}6y^5\,\mathrm dz\,\mathrm dy\,\mathrm dx$

$=\displaystyle6\int_{-1}^1\int_{-1}^1y^5(2-x^4-y^4)\,\mathrm dy\,\mathrm dx$

The integrand wrt $y$ is odd and symmetric about $y=0$, so the remaining integrals vanish and the flux is 0.