 # 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. necessaryh 2020-12-06 Answered
Use the Divergence Theorem to calculate the surface integral $\int {\int }_{S}F·dS$, that is, calculate the flux of F across S.
$F\left(x,y,z\right)=\left(\mathrm{cos}\left(z\right)+x{y}^{2}\right)i+x{e}^{-z}j+\left(\mathrm{sin}\left(y\right)+{x}^{2}z\right)k$
S is the surface of the solid bounded by the paraboloid $z={x}^{2}+{y}^{2}$ and the plane z = 9.
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Step 1
Here,
$F=<\mathrm{cos}z+x{y}^{2},x{e}^{-z},\mathrm{sin}y+{x}^{2}z>$
Step 2
$\mathrm{\nabla }F=\frac{\partial Fx}{\partial x}+\frac{\partial Fy}{\partial y}+\frac{\partial Fz}{\partial z}$
$={y}^{2}+0+{x}^{2}$
$={x}^{2}+{y}^{2}$
Step 3
Assume, $\left(x,y,z\right)=\left(r\mathrm{cos}0,r\mathrm{sin}0,z\right)$ where
0 $0<0<2\pi$
$0
Apply divergence theorem.
$\int {\int }_{S}Fds=\int \int {\int }_{E}\mathrm{\nabla }FdV$
$=\int \int {\int }_{E}\left({x}^{2}+{y}^{2}\right)dV$
$={\int }_{0}^{9}{\int }_{0}^{2\pi }{\int }_{0}^{\sqrt{z}}\left({r}^{2}{\mathrm{cos}}^{2}0+{r}^{2}{\mathrm{sin}}^{2}0\right)rdrd0dz$
$={\int }_{0}^{9}{\int }_{0}^{2\pi }{\int }_{0}^{\sqrt{z}}{r}^{3}drd0dz$
$={\int }_{0}^{9}{\int }_{0}^{2\pi }{\left[\frac{{r}^{4}}{4}\right]}_{0}^{\sqrt{z}}d0dz$
$={\int }_{0}^{9}{\int }_{0}^{2\pi }\left[\frac{\sqrt{{z}^{4}}}{4}-0\right]d0dz$
$=\frac{1}{4}{\int }_{0}^{9}{z}^{2}{\left[0\right]}_{0}^{2\pi }dz$
$=\frac{1}{4}{\int }_{0}^{9}{z}^{2}\left[2\pi -0\right]dz$
$=\frac{2\pi }{4}{\int }_{0}^{9}{z}^{2}dz$
$=\frac{\pi }{2}\left[\frac{{9}^{3}}{3}-0\right]$
$=\frac{243\pi }{2}$
Step 4
Result: