Use the Divergence Theorem to calculate the surface integral int int_S F · dS, that is, calculate the flux of F across S. F(x,y,z) = (cos(z)+xy^2) i + xe^-z j + (sin(y)+x^2z) k S is the surface of the solid bounded by the paraboloid z = x^2+y^2 and the plane z = 9.

Step 1
Here,
${\displaystyle F=<\cos z+x{y}^2,x{e}^{-z},\sin y+x^{2}z>}$
Step 2
Find grad F.
${\displaystyle \mathrm{\nabla }F=\frac{\partial Fx}{\partial x}+\frac{\partial Fy}{\partial y}+\frac{\partial Fz}{\partial z}}$
${\displaystyle =y^2+0+x^2}$
${\displaystyle =x^2+y^2}$
Step 3
Assume, ${\displaystyle (x,y,z)=(r\cos 0,r\sin 0,z)}$ where
0 ${\displaystyle 0<0<2\pi }$
${\displaystyle 0<r<\sqrt{z}}$
Apply divergence theorem.
${\displaystyle \int {\int }_{S}Fds=\int \int {\int }_E\mathrm{\nabla }FdV}$
${\displaystyle =\int \int {\int }_E(x^2+y^2)dV}$
${\displaystyle ={\int }_0^9{\int }_0^{2\pi }{\int }_0^{\sqrt{z}}(r^2{\cos }^{2}0+r^2{\sin }^{2}0)rdrd0dz}$
${\displaystyle ={\int }_0^9{\int }_0^{2\pi }{\int }_0^{\sqrt{z}}{r}^{3}drd0dz}$
${\displaystyle ={\int }_0^9{\int }_0^{2\pi }{\left[\frac{r^4}{4}\right]}_0^{\sqrt{z}}d0dz}$
${\displaystyle ={\int }_0^9{\int }_0^{2\pi }[\frac{\sqrt{z^4}}{4}-0]d0dz}$
${\displaystyle =\frac{1}{4}{\int }_0^9{z}^2{\left[0\right]}_0^{2\pi }dz}$
${\displaystyle =\frac{1}{4}{\int }_0^9{z}^2[2\pi -0]dz}$
${\displaystyle =\frac{2\pi }{4}{\int }_0^9{z}^{2}dz}$
${\displaystyle =\frac{\pi }{2}[\frac{9^3}{3}-0]}$
${\displaystyle =\frac{243\pi }{2}}$
Step 4
Result:

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