# Using the Divergence Theorem, evaluate int int_S F.NdS, where F(x,y,z)=(z^3i-x^3j+y^3k) and S is the sphere x^2+y^2+z^2=a^2, with outward unit normal vector N. Question
Green's, Stokes', and the divergence theorem Using the Divergence Theorem, evaluate $$\displaystyle\int\int_{{S}}{F}.{N}{d}{S}$$, where $$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({z}^{{3}}{i}-{x}^{{3}}{j}+{y}^{{3}}{k}\right)}$$ and S is the sphere $$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={a}^{{2}}$$, with outward unit normal vector N. 2021-01-20
Step 1
Divergence theorem relates surface integrals and volume integrals. By using the Gauss divergence theorem we can evaluate this surface integral.
The Gauss divergence theorem formula can be stated as follows:
$$\displaystyle\int\int_{{S}}{F}.{N}{d}{S}=\int\int\int_{{V}}\div{F}.{d}{V}$$
$$\displaystyle\div{F}={\left(\partial\frac{{{z}^{{3}}{i}}}{{\partial{x}}}\right)}+{\left(\partial\frac{{-{x}^{{3}}{j}}}{{\partial{y}}}\right)}+{\left(\partial\frac{{{y}^{{3}}{k}}}{{\partial{z}}}\right)}$$
Step 2
div F = 0, since the partial derivative of $$\displaystyle{z}^{{3}}$$ with respect to x is zero and partial derivative of $$\displaystyle-{x}^{{3}}$$ with respect to y is zero and the partial derivative of $$\displaystyle{y}^{{3}}$$ with respect to z is zero.
Then $$\displaystyle\int\int\int_{{V}}{0}.{d}{V}={0}$$
Since $$\displaystyle\int\int_{{S}}{F}.{N}{d}{S}=\int\int\int_{{V}}\div{F}.{d}{V}$$
Then by Gauss Divergence theorem, we can say $$\displaystyle\int\int_{{S}}{F}.{N}{d}{S}={0}$$.
Step 3
So the final answer is zero by the Gauss Divergence theorem.

### Relevant Questions Use the divergence theorem to evaluate $$\displaystyle\int\int_{{S}}{F}\cdot{N}{d}{S}$$, where $$\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}$$ and S is the boundary of the cube defined by $$\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}$$. 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. Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem $$\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
where n(t) is the outward unit normal vector to C. Let $$\displaystyle{F}={\left[{x}^{{2}},{0},{z}^{{2}}\right]}$$, and S the surface of the box $$\displaystyle{\left|{{x}}\right|}\le{1},{\left|{{y}}\right|}\le{3},{0}\le{z}\le{2}$$.
Evaluate the surface integral $$\displaystyle\int\int_{{S}}{F}\cdot{n}{d}{A}$$ by the divergence theorem. 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. 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. Use Stokes' Theorem to evaluate $$\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$,
S is the part of the cone $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis. z = x Let be the curve of intersection of the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the plane , oriented positively when viewed from above . Let S be the inside of this curve , oriented with upward -pointing normal . Use Stokes ' Theorem to evaluate $$\displaystyle\int{S}{c}{u}{r}{l}{F}\cdot{d}{S}{\quad\text{if}\quad}{F}={y}{i}+{z}{j}+{2}{x}{k}$$. $$\displaystyle{F}={\left\langle{z}-{x},{x}-{y},{2}{y}-{z}\right\rangle}$$, D is the region between the spheres of radius 2 and 4 centered at the origin. $$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
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.