# Use the Divergence Theorem to find the flux of F = xy^2i + x^2yj + yk outward through the surface of the region enclosed by the cylinder x^2 + y^2 = 1 and the planes z = 1 and z =-1.

Question
Use the Divergence Theorem to find the flux of $$\displaystyle{F}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}$$ outward through the surface of the region enclosed by the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the planes z = 1 and z =-1.

2020-12-30
Step 1
Divergence Theorem.
Let F be a vector field whose components have continuous first partial derivatives and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface's outward unit normal field n equals the triple integral of the divergence grad * F over the solid region D enclosed by the surface.
$$\displaystyle\int\int_{{S}}{F}\cdot{n}{d}\sigma=\int\int_{{D}}\int\nabla\cdot{F}{d}{V}$$.
Step 2
Write the vector field.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}$$
Find the divergence of Equation(1).
$$\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{y}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{z}}}{\left({y}\right)}$$
$$\displaystyle={\left({y}^{{2}}\right)}+{\left({x}^{{2}}\right)}+{\left({0}\right)}$$
$$\displaystyle={x}^{{2}}+{y}^{{2}}$$
Step 3
Find the flux of the vector field.
Flux = $$\displaystyle\int\int_{{D}}\int\nabla{F}{d}{V}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{\int_{{-{{1}}}}^{{1}}}{r}^{{2}}{\left.{d}{z}\right.}{r}{d}{r}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{{\left[{z}\right]}_{{1}}^{{1}}}{d}{r}{\left.{d}{z}\right.}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{\left({1}+{1}\right)}{d}{r}{\left.{d}{z}\right.}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{2}{r}^{{3}}{d}{r}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{{2}{r}^{{4}}}}{{4}}\right]}_{{0}}^{{1}}}{d}{0}$$
Flux = $$\displaystyle{\int_{{0}}^{{{2}\pi}}}{\left[{\left(\frac{{{\left({1}\right)}^{{4}}}}{{2}}\right)}-{0}\right]}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}\frac{{1}}{{2}}{d}{0}$$
$$\displaystyle=\frac{{1}}{{2}}{{\left[{0}\right]}_{{0}}^{{{2}\pi}}}$$
$$\displaystyle=\frac{{1}}{{2}}{\left({2}\pi-{0}\right)}$$
$$\displaystyle=\pi$$
Step 4
Therefore, the outward flux of the vector field $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}$$ by using the Divergence Theorem is $$\displaystyle\pi$$.

### Relevant Questions

Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
S is the surface of the solid bounded by the cylinder $$\displaystyle{y}^{{2}}+{z}^{{2}}={9}$$ and the planes x = −3 and x = 1.
Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Divergence Theorem for more general regions Use the Divergence Theorem to compute the net outward flux of the following vector fields across the boundary of the given regions D.
$$\displaystyle{F}={\left\langle{z}-{x},{x}-{y},{2}{y}-{z}\right\rangle}$$, D is the region between the spheres of radius 2 and 4 centered at the origin.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Set up the integral for the divergence theorem both ways. Then find the flux.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}$$
E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. $$\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}$$
Consider the vector field $$\displaystyle{F}={\left\langle{5}{z},{x},{5}{y}\right\rangle}$$ and the surface which is the part of the elliptic paraboloid $$\displaystyle{z}={x}^{{2}}+{5}{y}^{{2}}$$ that lies below the plane z = 5. Calculate curl(F) and then apply Stokes' Theorem to compute the exact magnitude of the flux of curl(F) through the surface using line integral. You do not need to cinfirm your answer by evaluating the double integral of curl(F) over the surface(the right-hand side of Stokes' Theorem).
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.