MATH 251 Lecture 35

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$C$ is a path through space, $R$ is a 3D region in space, $\Omega$ represents the boundary surface of the enclosed region, and $D$ represents the domain of $\Omega$.

Vector Analysis Theorems

1. FTC: $\int _{a}^{b}f'\left(x\right)\,\mathrm {d} x=f\left(b\right)-f\left(a\right)$
2. FTLI: $\int _{C}{\vec {\nabla }}f\cdot \mathrm {d} {\vec {r}}=f\left(Q\right)-f\left(P\right)$
3. Green's Theorem: $\iint _{D}\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,\mathrm {d} A=\oint _{C}P\,\mathrm {d} x+Q\,\mathrm {d} y$
4. Stokes' Theorem: $\iint _{\Omega }{\vec {\nabla }}\times {\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S=\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {r}}$
5. Divergence Theorem: $\iiint _{R}{\vec {\nabla }}\cdot {\vec {F}}\,\mathrm {d} V=\iint _{\Omega }{\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S$

All have something in common: Integrating over the boundary (rhs) means something about integrating over the interior:

$\int _{R}\mathrm {d} \omega =\int _{\partial R}\omega$

Divergence Theorem

(See Vector Analysis Theorems#Gauss' (Divergence) Theorem→)

Example

Given ${\vec {F}}=3\,x^{2}\,{\hat {\imath }}-2\,y\,{\hat {\jmath }}+\sin {z}\,{\hat {k}}$, find its flux over the sphere of radius 3 centered at the origin:

$\Phi =\iint _{\Omega }{\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S$.

We could plug and chug this integral, but the divergence theorem gives us another option:

${\vec {\nabla }}\cdot {\vec {F}}=6x-2+\cos {z}$, and therefore,

$\Phi =\iiint _{R}6x-2+\cos {z}\,\mathrm {d} V$ such that $R=\left\{(x,y,z)\mid x^{2}+y^{2}+z^{2}\leq 9\right\}$

1. $\iiint _{R}6x\,\mathrm {d} V=0$ since it's the average value of $x$ over a symmetric region.
2. $-2\iiint _{R}\,\mathrm {d} V=-72\pi$ since it's the volume of the region $V={\tfrac {4}{3}}\pi r^{3}$
3. $\iiint _{R}\cos {z}\,\mathrm {d} V=\ldots$ (let's not evaluate this integral)

Another Example

Let ${\vec {F}}=\left(y^{3}-2\,z\right)\,{\hat {\imath }}+\left(3\,x\,y^{2}\right)\,{\hat {\jmath }}+{\hat {k}}$.

Find $\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {r}}$, where $C$ represents the curve $x^{2}+y^{2}=1$, $z=0$ counter-clockwise when viewed from above.

We could parameterize this with $x=\cos {t},\,y=\sin {t}$, but I wouldn't want to integrate $3\cos {t}\sin ^{2}{t}$.

According to Stokes' theorem, the integral of a curve is related to the integral of the curl over any surface bounded by that curve.

${\vec {\nabla }}\times {\vec {F}}=-2{\hat {\jmath }}$.

We know that ${\vec {F}}\neq {\vec {\nabla }}f$, but ${\vec {F}}={\vec {\nabla }}(xy^{3}+z)-2z{\hat {\imath }}$. Taking the curl wipes out the conservative parts of a vector field.

Now we have $\iint _{\Omega }-2{\hat {\jmath }}\cdot {\hat {n}}\,\mathrm {d} S$, where $\Omega$ represents the simplest surface bounded by $C$: a unit disk in the $xy$ plane.

${\vec {X}}(u,v)=\left\langle u,v,0\right\rangle$, but we can easily tell that ${\hat {n}}={\hat {k}}$ (follows RHR since the curve is going counter-clockwise).

$\iint _{\Omega }-2{\hat {\jmath }}\cdot {\hat {k}}\mathrm {d} A=0$

Yet Another Example

Calculate $\iint _{\Omega }{\vec {\nabla }}\times {\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S$ over the upper hemisphere $x^{2}+y^{2}+z^{2}=1$, $z\geq 0$, where ${\hat {n}}$ points radially outward.

${\vec {F}}=(y^{3}-2z){\hat {\imath }}+(3xy^{2}){\hat {\jmath }}+{\hat {k}}$

According to Stokes' theorem, $\iint _{\Omega }{\vec {\nabla }}\times {\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S=\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {r}}=0$, where $\Omega$ is any other surface with same boundary curve $C$.