Course Blog

Math 275 – May 10 (Final Notes)

May 10, 2009 by ketchers

Here is a printable version of this post.

All of the points mentioned below are covered in your book and in the online class notes – with examples. Of course everything was covered with examples in class as well. You have plenty of homework assigned so I will not suggest more here. for examples look at the other online notes and in the text.

1. Line integrals

Definition of path, curve, simple path/curve, closed path/curve, Jordan path/curve, smooth curve, and piecewise smooth curve.
You need to know what ${\int_C f\cdot \vec T\,ds}$ and ${\int_C f\cdot\vec n\,ds}$ mean and how to calculate these.
You should know how to use line integrals to calculate mass, average value of ${f}$ along ${C}$ , center of mass, moments of inertia, etc.
You need to understand the notion of orientation of a curve.
You need to understand and know how to calculate, flux through a curve, flow along a curve, and work along a curve.
You need to know what circulation is and what the notation ${\oint_C f\,ds}$ means. You need to know what the standard (outward) orientation of a Jordan curve is.

2. Del

You should know what ${\nabla}$ is and what ${\nabla\cdot F}$ (divergence of ${F}$ ), ${\nabla \times F}$ (curl of ${F}$ ) are.
You should know some trivial facts: ${\nabla\times(\nabla f)=0}$ (since this is just ${(\nabla\times\nabla)f}$ ) so in particular the curl of the gradient is always ${0}$ . ${\nabla\cdot(\nabla\times F)=0}$ – recall ${\vec v\cdot(\vec u\vec w)}$ is the triple product and is the same as ${\det(\vec u,\vec v,\vec w)}$ , so ${\det(\nabla,\nabla,F)=0}$ since the determinant of a matrix with two identical rows is ${0}$ . So the divergence of the curl is always ${0}$ .

3. Conservative fields

Definition of conservative field as well as restrictions on the domain so that the definition makes sense, i.e., open and path connected.
Path independence, equivalence of path independence with conservative.
Gradient fields/potential functions, equivalence with conservative fields under certain assumptions.
Fundamental theorem for line integrals, ${\int_C F\,dr=f(B)-f(A)}$ where ${F}$ is a conservative field defined on an open path connected region ${R}$ , ${C}$ is a piecewise smooth curve from ${A}$ to ${B}$ lying in ${R}$ , and ${F=\nabla f}$ for some potential function ${f}$ .
Test to check when a field is conservative, including conditions on when the test works. (Open, path connected, simply connected region; ${F}$ has continuous second partials.)
Be able to find a potential function for a conservative field and use this to evaluate a line integral.

4. Surface integrals

You need to know what a smooth surface is and how they fit together to form piecewise smooth surfaces.
You need to know and understand what an orientation of a smooth/piecewise smooth surface is and how to find an orientation if it exists.
You need to know what surface integrals are and how to compute them. In particular you need to know what ${\iint_S f\,d\sigma}$ means, how to interpret ${d\sigma}$ – in the notes there are several ways of interpreting ${d\sigma}$ mentioned.
You should know how to apply surface integrals to find surface area, mass of a thin surface, center of mass of a thin surface, moment of inertia of a thin surface, average value of ${f}$ on ${S}$ , etc.
You should know the definition of flux of ${F}$ through ${S}$ and how to compute this.

5. The fundamental theorems of calculus – Green’s, Stoke’s, Divergence

You should know and be able to use the normal and tangential forms of Green’s theorem in the plane. In particular you must know the hypotheses of Green’s theorem and when it applies. You must know what the induced orientation of the boundary of ${R}$ is and how to compute it.
You should be able to use Green’s to simplify the calculation of certain line integrals and conversely to compute area of planar regions.
You should have some idea of how Green’s gives the test for path independence and why simple connectedness is important.
You should understand the way Green’s theorem can be used in the case that ${\nabla\cdot F=0}$ or ${(\nabla\times F)\cdot \hat k=0}$ . For example given a field ${F}$ with ${0}$ divergence on a region ${R}$ and a Jordan curve ${C}$ in ${R}$ , then ${\oint_C F\cdot\vec n\,ds=\oint_{C'} F\cdot\vec n\,ds}$ where ${C'}$ might be taken to be a much simpler curve. (This might be the point of the problem, it is up to you to pick a simple ${C'}$ and evaluate the line integral there.)
Stoke’s theorem is essentially just the tangential form of Green’s theorem for arbitrary surfaces satisfying certain conditions – that you must know.
You should be able to use Stoke’s theorem to calculate the flux of ${\nabla\times F}$ through some surface via computing a simple line integral on the boundary.
You should know why you can replace a given surface by a much simpler one having the same boundary and use this in computations of ${\iint_S\nabla\times F\,\vec n\,d\sigma}$ . For example if I ask you to compute the flux of ${\nabla\times F}$ through the parabolic surface ${S}$ ${z=x^2+y^2}$ from ${z=0}$ to ${z=2}$ . Then you know ${\int_S \nabla\times F\vec n\,ds=\int_{S'}\nabla\times F\vec n\,ds}$ where ${S'}$ is just the unit disk centered on the ${z}$ -axis sitting at ${z=2}$ . (This is also a consequence of the divergence theorem.) Conversely, Stoke’s can be use to calculate a line integral ${\oint_C F\cdot dr}$ via a corresponding surface integral ${\iint_S \nabla \times F\cdot\vec n\,d\sigma}$ where ${\partial S=C}$ .
The divergence theorem is just the three dimensional version of the normal form of Green’s theorem.
One important case (just as with Green’s and Stoke’s) is that when ${\nabla\cdot F=0}$ ( ${0}$ -divergence), then the flux integral through a surface is “independent of the surface” in some sense.
You should know the relevance of fields satisfying ${F(\vec r)=\frac{k}{|\vec r|^n}\hat r}$ (radial fields that are inversely proportional to the ${n^{\text{th}}}$ power of the distance from the origin and in particular that the divergence for these fields is ${0}$ when ${n=2}$ ).

Not to under emphasise this: Know the main theorems and definitions, know the conditions under which the theorems apply – i.e. the hypotheses of the theorems.

Posted in Math 275, Spring 2009 | Leave a Comment »

Math 275 – May 5

May 3, 2009 by ketchers

Here is a printable version of this post.

1. Divergence Theorem

The divergence theorem is also known as Green’s theorem (for three dimensions) and as Gauss’ theorem.

Theorem 1 Let ${D}$ be a closed bounded region in ${{\mathbb R}^3}$ with ${\partial D}$ being the union of finitely many disjoint closed smooth surfaces. The boundary of ${D}$ is oriented with orientation ${\vec n}$ so that the ${\vec n}$ points into ${D}$ . Let ${F:{\mathbb R}^3\rightarrow{\mathbb R}^3}$ be a vector field defined on an open region ${R}$ including ${D}$ and which is continuously differentiable in ${R}$ . Then
$\displaystyle \iiint_D \nabla\cdot F\,dV=\iint_{\partial D} F\cdot \vec n\,d\sigma$

Notice that this really is Green’s theorem pushed up to 3 dimensions and in fact there is a general theorem that could just be called the “Fundamental Theorem of Calculus” that would cover all these (the usual FTC is just Green’s theorem in one dimension!) In the following take ${n=1}$ to get FTC from Calc I, ${n=2}$ gives Green’s Theorem (and hence Stoke’s Theorem), ${n=3}$ gives the Divergence Theorem.

Theorem 2 Let ${D}$ be a closed bounded region in ${{\mathbb R}^n}$ with ${\partial D}$ being the union of finitely many disjoint closed smooth ${n-1}$ -dimensional regions. The boundary of ${D}$ is oriented with orientation ${\vec n}$ so that the ${\vec n}$ points into ${D}$ . Let ${F:{\mathbb R}^n\rightarrow{\mathbb R}^n}$ be a vector field defined on an open region ${R}$ including ${D}$ and which is continuously differentiable in ${R}$ . Then
$\displaystyle \int_D \nabla \cdot F\,dV=\oint_{\partial D} F\cdot \vec n\,d\sigma$

The proof will be very much like that of Green’s theorem. First we will prove the theorem for particularly simple regions and then argue geometrically to extend the result to more general regions. The simple regions are called projectable.

Let ${D}$ be a solid and ${D_{xy}}$ be the projection (shadow) of ${D}$ on the ${xy}$ -plane. If there are are smooth ${z_1,z_2:D_{xy}\rightarrow\partial D}$ so that ${z_1(x,y)\le z_2(x,y)}$ for all ${(x,y)\in D_{xy}}$ such that ${D}$ can be viewed as the solid bounded by

$\displaystyle S_1=\{(x,y,z_1(x,y)):(x,y)\in D_{xy}\}$

$\displaystyle S_2=\{(x,y,z_2(x,y)):(x,y)\in D_{xy}\}$

$\displaystyle S_3=\{(x,y,z):(x,y)\in \partial D_{xy} \text{ and }z_1(x,y) \le z \le z_2(x,y)\}$

then ${D}$ is called ${xy}$ -projectable.

Similarly ${xz}$ -projectable and ${yz}$ -projectable are defined and ${D}$ is projectable if it is ${xy}$ , ${xz}$ , and ${yz}$ -projectable.

Proof: (Proof sketch for the divergence theorem) Let ${D}$ be projectable and let the outward normal ${\vec n=\langle n_1,n_2,n_3}$ , we want to show

$\displaystyle \iint_{\partial D} F\cdot \vec n\,d\sigma=\iiint_D \nabla\cdot F\,dV$

Expanding this it becomes

$\displaystyle \iint_S Mn_1+Nn_2+Pn_3\,d\sigma= \iiint_D \frac{\partial M}{\partial x}+ \frac{\partial N}{\partial y}+ \frac{\partial P}{\partial z}\,dV$

so it suffices to show

$\displaystyle \iint_S Mn_1\,d\sigma=\iiint_D \frac{\partial M}{\partial x}\,dV$

$\displaystyle \iint_S Nn_2\,d\sigma=\iiint_D \frac{\partial N}{\partial y}\,dV$

$\displaystyle \iint_S Pn_3\,d\sigma=\iiint_D \frac{\partial P}{\partial z}\,dV$

Suppose ${D}$ is ${xy}$ -projectable and we show

$\displaystyle \iint_S Pn_3\,d\sigma=\iiint_D \frac{\partial P}{\partial z}\,dV$

First calculate ${\iiint_D\frac{\partial P}{\partial z}\,dV}$ using Fubini.

$\displaystyle \iiint_D\frac{\partial P}{\partial z}\,dV= \iint_{D_{xy}}\int_{z_1(x,y)}^{z_2(x,y)}\frac{\partial P}{\partial z}\,dz\,dA$

$\displaystyle =\iint_{D_{xy}} \left[P(x,y,z_2(x,y))-P(x,y,z_1(x,y))\right]\,dA$

${\iint_{\partial D}Pn_3\,d\sigma= \sum_{i=1}^3 \iint_{S_i}Pn_3\,d\sigma}$ so we must compute ${\iint_{S_i}Pn_3\,d\sigma}$ for ${i=1,2,3}$ . For ${i=3}$ , ${n_3=0}$ so ${\iint_{S_3}Pn_3\,d\sigma=0}$ . For ${i=1,2}$ , define ${s_i:D_{xy}\rightarrow S_i}$ by ${s_i(x,y)=(x,y,z_i(x,y))}$ .

For ${i=1}$ , the norm must point outward, this is reverse of the standard orientation:

$\displaystyle (r_1)_y\times(r_1)_x = \langle 0,1,(z_1)_y \rangle \times \langle 1,0,(z_1)_x \rangle =\langle (z_1)_x,(z_1)_y,-1 \rangle$

and ${\vec n=\frac{(r_1)_y\times(r_1)_x}{|(r_1)_y\times(r_1)_x|}}$ and ${d\sigma=|(r_1)_y\times(r_1)_x|\,dA}$ so ${n_3\,d\sigma=-dA}$

$\displaystyle \iint_{S_1}Pn_3\,d\sigma= \iint_{D_{xy}}-P(x,y,z_1(x,y))\,dA$

For ${i=2}$ ,

$\displaystyle (r_2)_x\times(r_2)_y= \langle 1,0,(z_2)_x \rangle \times \langle 0,1,(z_2)_y \rangle =\langle -(z_2)_x,-(z_2)_y,1 \rangle$

and ${n_3\,d\sigma=dA}$ (this is like the calculation for ${i=1}$ )

$\displaystyle \iint_{S_2}Pn_3,d\sigma= \iint_{D_{xy}}P(x,y,z_2(x,y))\,dA$

So we have

$\displaystyle \iint_{\partial D}Pn_3\,d\sigma= \iint_{D_{xy}}P(x,y,z_2(x,y))\,dA+\iint_{D_{xy}}-P(x,y,z_1(x,y))\,dA$

$\displaystyle =\iint_{D_{xy}}[P(x,y,z_2(x,y)-P(x,y,z_1(x,y)]\,dA$

and hence

$\displaystyle \iint_S Pn_3\,d\sigma=\iiint_D \frac{\partial P}{\partial z}\,dV$

which is what we wanted. The other cases are proved in a similar fashion.

That every region as in the hypothesis of the divergence theorem can be sufficiently approximated by projectable regions is left to the imagination of the reader. $\Box$

The divergence theorem itself provides a reason for calling ${\nabla\cdot F}$ the divergence. Let ${D_r(P)}$ be a sphere centered at ${P}$ of radius ${r}$ . The mean value theorem for multiple integrals says that for some ${P^*\in D_r(P)}$ , ${\iiint_{D_r(P)} f\,dV=f(P^*)V_r}$ where ${V_r}$ is the volume of a sphere of radius ${r}$ . So we have

$\displaystyle \iint_{\partial D_r(P)} F\cdot \vec n\,d\sigma=\nabla\cdot F(P^*)V_r$

$\displaystyle \nabla\cdot F(P)=\lim_{r\rightarrow0^+}\frac{1}{V_r}\iint_{\partial D_r(P)} F\cdot \vec n\,d\sigma$

That is

$\displaystyle \nabla \cdot F(P)=\lim_{r\rightarrow0^+}\frac{\text{flux of }F\text{ outward through the sphere of radius }r\text{ centered at }P}{\text{volume of the sphere of radius }r\text{ centered at }P}$

this is the flux density, i.e., flux/unit of volume at a point, and that is an interpretation of divergence.

Example 1. Compute the flux of ${F(x,y,z)=\langle xy,y^2,z\rangle}$ outward through thesurface of the solid cone ${z=2\sqrt{x^2+y^2}}$ for ${0\le z\le 2}$ first just directly using surface integrals and then using the divergence theorem.

I will call the cone ${D}$ the boundary of ${D}$ consists of two surfaces ${S_1}$ is the surface of the cone and ${S_2}$ is the top. I will use polar coordinates for the parameterization, so ${z=2\sqrt{x^2+y^2}=2r}$ , (this is not the only way.) In each case ${R=[0,1]\times[0,2\pi]}$ and

$\displaystyle s_1:R\rightarrow S_1\text{ is given by }(r,\theta)\mapsto \langle r\cos(\theta),r\sin(\theta),2r\rangle$

$\displaystyle s_2:R\rightarrow S_1\text{ is given by }(r,\theta)\mapsto \langle r\cos(\theta),r\sin(\theta),2\rangle$

A little thought gives

$\displaystyle \vec n_1\,d\sigma = -[(s_1)_r\times (s_1)_\theta]\,dr\,d\theta$

$\displaystyle \vec n_2\,d\sigma = [(s_2)_r\times (s_2)_\theta]\,dr\,d\theta$

and little calculation gives

$\displaystyle -[(s_1)_r\times (s_1)_\theta]= \langle 2r\cos(\theta),2r\sin(\theta),-r\rangle$

$\displaystyle [(s_2)_r\times (s_2)_\theta]=\langle 0,0,r\rangle$

$\displaystyle \text{flux outward through }S_1=\iint_{S_1} F\cdot\vec n_1\,d\sigma$

$\displaystyle =\int_0^{2\pi}\int_0^1 \langle r^2\cos(\theta)\sin(\theta),r^2\sin^2(\theta),2r\rangle \cdot \langle 2r\cos(\theta),2r\sin(\theta),-r\rangle\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi}\int_0^1 2r^3\cos^2(\theta)\sin(\theta)+2r^3\sin^3(\theta)-2r^2\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi} \frac{1}{2}\cos^2(\theta)\sin(\theta)+\frac{1}{2}\sin^3(\theta)-\frac{2}{3}\,d\theta$

Since ${\sin^3(\theta)=(1-\cos^2(\theta))\sin(\theta)}$ this becomes

$\displaystyle \int^{2\pi}_0 \frac{\sin(\theta)}{2}-\frac{2}{3}\,d\theta=\frac{-4\pi}{3}$

$\displaystyle \text{flux outward through }S_2=\iint_{S_2} F\cdot\vec n_2\,d\sigma$

$\displaystyle =\int_0^{2\pi}\int_0^1 \langle r^2\cos(\theta)\sin(\theta),r^2\sin^2(\theta),2\rangle \cdot\langle 0,0,r\rangle\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi}\int_0^12r\,dr\,d\theta=2\pi$

$\displaystyle \text{flux outward through }\partial D=2\pi-\frac{4\pi}{3}=\frac{2\pi}{3}$

Now using the divergence theorem

$\displaystyle \iiint_D \nabla\cdot F\,dV=\iiint 3y+1\,dV$

Using cylindrical coordinates this is

$\displaystyle \int_0^{2\pi}\int_0^1\int_{2r}^2 [3r\sin(\theta)+1]r\,dz\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi}\int_0^1[3r^2\sin(\theta)+r] z|^2_{2r} = \int_0^{2\pi}\int_0^1[3r^2\sin(\theta)+r](2-2r)\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi}\int_0^1 6(r^2-r^3)\sin(\theta)- 2(r-r^2)\,dr\,d\theta$

$\displaystyle =\int_0^{2\pi} 6(1/3-1/4)\sin(\theta)+2(1/2-1/3)\,d\theta$

$\displaystyle =\int_0^{2\pi} \frac{\sin(\theta)}{2}+\frac{1}{3}\,d\theta=\frac{2\pi}{3}$

Example 2. One useful consequence of the Divergence theorem is that if ${\nabla\cdot F=0}$ on an open region ${D}$ of ${{\mathbb R}^3}$ and ${D_1\subseteq D_2\subseteq D}$ are solids with ${\partial D_1\cap \partial D_2=\emptyset}$ , then

$\displaystyle \iint_{\partial D_1} f\cdot \vec n_1\,d\sigma=\iint_{\partial D_2} f\cdot \vec n_2\,d\sigma$

For example the electric field produced by a point charge ${q}$ at the origin is

$\displaystyle E(x,y,z)=\frac{q}{4\pi\epsilon}\frac{\langle x,y,z\rangle}{|\langle x,y,z\rangle|^3}$

See \href{http://en.wikipedia.org/wiki/Coulomb for more on this.

A little calculation shows ${\nabla\cdot E = 0}$ for ${\langle x,y,z\rangle\ne\langle0,0,0\rangle}$ . The strength of the field passing through the surface ${S}$ of a solid containing the origin is

$\displaystyle \text{flux}_S=\iint_S E\cdot \vec n\,d\sigma$

For complicated ${S}$ this might be difficult to compute, but we can find ${\delta>0}$ so that the ball of radius ${\delta}$ centered at the origin, i.e., ${B_\delta}$ , is completely contained within the solid bounded by ${S}$ . The boundary of ${B_\delta}$ , ${\partial B_\delta=S_\delta}$ , where ${S_\delta}$ is the sphere of radius ${\delta}$ centered at the origin. The divergence theorem gives

$\displaystyle \text{flux through }S=\text{flux through } S_\delta =\iint_{S_\delta}E\cdot \vec n\,d\sigma$

This is not hard to calculate!

Parameterize ${S_\delta}$ by

$\displaystyle s:[0,\pi]\times[0,2\pi]\rightarrow S_\delta\text{ where }(\phi,\theta)\mapsto \langle \delta\sin(\phi)\cos(\theta), \delta\sin(\phi)\sin(\theta),\delta\cos(\phi)\rangle$

A little calculation gives that ${s_\phi\times s_\theta}$ points outwards and

$\displaystyle s_\phi\times s_\theta=\delta\sin(\phi)s$

$\displaystyle \text{outward flux through}S_\delta= \iint_{S_\delta}E\cdot(s_\phi\times s_\theta)\,d\theta\,d\phi$

$\displaystyle =\frac{q}{4\pi\epsilon\delta^3}\int_0^\pi\int_0^{2\pi}\delta\sin(\phi)|s|^2\,d\theta\,d\phi= \frac{q}{4\pi\epsilon\delta^3}\int_0^\pi\int_0^{2\pi} \delta^3 \sin(\phi) d\theta\,d\phi$

$\displaystyle =\frac{q}{4\pi\epsilon}\int_0^\pi2\pi\sin(\phi)\,d\phi= \left.\frac{q}{2\epsilon}\left(-\cos(\phi)\right)\right|_0^\pi=\frac{q}{\epsilon}$

So for any surface of a solid containing the origin

$\displaystyle \text{flux}_S=\frac{q}{\epsilon}$

Example 3. The Divergence Theorem can be used to find volume just as Green’s Theorem could be used to find volume. Given a solid ${D}$ we can take any field ${F}$ with ${\nabla\cdot F=1}$ and get

$\displaystyle \text{volume of }D=\iiint_D\,dV=\iint_{\partial D}F\cdot\vec n\,d\sigma$

Consider the solid, ${D}$ , formed by rotating the curve ${x=\cos(u)}$ , ${z=\sin(2u)}$ for ${-\pi/2\le u\le \pi/2}$ about the ${z}$ -axis. Take ${F(x,y,x)=\langle 0,0,z \rangle}$ so ${\nabla\cdot F=1}$ . The boundary of ${D}$ , (a surface of rotation) is parameterized by ${s:[-\pi/2,\pi/2]\times[0,2\pi]\rightarrow \partial D}$ by ${s(u,v)=\langle \cos(u)\cos(v), \cos(u)\sin(v),\sin(2u) \rangle}$ . A moments thought reveals the outward normal to be given by

$\displaystyle -(s_u\times s_v)=\langle 2\cos(2u)\cos(u)\cos(v),2\cos(2u)\cos(u)\sin(v),\sin(u)\cos(u)\rangle$

$\displaystyle -F\cdot (s_u\times s_v)=\sin(2u)\sin(u)\cos(u)=1/2\sin^2(2u)$

hence

$\displaystyle \iint_{\partial D} F\cdot \vec n\,d\sigma=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\sin^2(2u)\,dv\,du$

$\displaystyle =\pi\int^{\pi/2}_{-\pi/2}\sin^2(2u)\,du=\pi\int_{-\pi/2}^{\pi/2}\frac{1-\cos(4u)}{2}\,du=\frac{\pi}{2}\left.u\right|^{\pi/2}_{-\pi/2}=\frac{\pi^2}{2}$

Problems:
14.8: 7, 13, 15, 17, 21, 25, 31

Posted in Math 275, Spring 2009 | Leave a Comment »

Math 275 – May 4

May 3, 2009 by ketchers

Here is a printable version of this post.

1. Stoke’s Theorem

Green’s theorem generalizes from regions in ${{\mathbb R}^2}$ and ${F:{\mathbb R}^2\rightarrow{\mathbb R}^2}$ to regions in the ${xy}$ -plane viewed as a surface in ${R^3}$ and ${F:{\mathbb R}^3\rightarrow{\mathbb R}^3}$ in this form it essentially does not change

$\displaystyle \iint_R \nabla\times F\cdot \hat k\,dA=\int_{\partial R} F \cdot dr$

Now if we replace ${R}$ with an oriented surface in ${{\mathbb R}^3}$ with orientation ${\vec n}$ with boundary consisting of finitely many non-intersecting Jordan curves oriented counterclockwise with respect to ${\vec n}$ then this becomes

$\displaystyle \iint_R \nabla\times F\cdot \vec n\,d\sigma=\int_{\partial R} F \cdot dr$

and this is Stoke’s Theorem.

Theorem 1 (Stoke’s Theorem) Let ${S}$ be a piecewise smooth oriented surface with orientation ${\vec n}$ and with a boundary consisting of finitely many disjoint Jordan curves, then
$\displaystyle \iint_S \nabla \times F \cdot \vec n\, d\sigma =\oint_{\partial S}F\cdot dr$

where the orientation of the boundary is positive with respect to ${\vec n}$ .

Written slightly differently Stoke’s theorem looks like

$\displaystyle \oint_{\partial R}M\,dx+N\,dy+P\,dz= \iint_R \textstyle{\left\langle \left(\frac{\partial P}{\partial y}- \frac{\partial N}{\partial z}\right), \left(\frac{\partial M}{\partial z}- \frac{\partial P}{\partial x}\right), \left(\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}\right) \right\rangle\cdot\vec n} \, d\sigma$

Proof: Let ${s:D\rightarrow S}$ parameterize ${S}$ where ${D}$ is now a region in the ${(u,v)}$ -plane satisfying the hypotheses of Green’s Theorem. The proof essentially just parameterizes ${S}$ via a planar set ${D}$ and applies Green’s theorem in ${D}$ . The details get a bit involved!

Recall

$\displaystyle \frac{\partial(z,x)}{\partial(u,v)}=\det\left( \begin{array}{cc} \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v} \end{array} \right)$

and similarly for other combinations like ${\frac{\partial(y,z)}{\partial(v,u)}}$ , etc. Let

$\displaystyle J_1=\frac{\partial(y,z)}{\partial(u,v)}\quad J_2=\frac{\partial(z,x)}{\partial(u,v)}\quad J_3=\frac{\partial(x,y)}{\partial(u,v)}\quad$

In this notation we have

$\displaystyle \vec n=s_u \times s_v=\langle J_1,J_2,J_3 \rangle$

$\displaystyle \iint_S\nabla\times F\cdot \vec n\,d\sigma=\iint_D\nabla\times F\cdot\langle J_1,J_2,J_1\rangle\,dA$

and

Using the chain rule ${dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv}$

$\displaystyle \int_{\partial S} M\,dx=\int_{\partial D}M\frac{\partial x}{\partial u}du+M\frac{\partial x}{\partial v}dv =\int_{\partial D}\textstyle{\left\langle M\frac{\partial x}{\partial u}, M\frac{\partial x}{\partial v}\right\rangle\cdot\langle du,dv\rangle}$

$\displaystyle \iint_{\partial S}F\cdot dr=\int_{\partial D} \textstyle{\left\langle M\frac{\partial x}{\partial u}+ N\frac{\partial y}{\partial u}+ P\frac{\partial z}{\partial u}, M\frac{\partial x}{\partial v}+ N\frac{\partial y}{\partial v}+ P\frac{\partial z}{\partial v}\right\rangle\cdot\langle du,dv\rangle}$

So we aim to show

$\displaystyle \iint_D\nabla\times F\cdot\langle J_1,J_2,J_1\rangle\,dA=$

$\displaystyle \iint_D\frac{\partial M}{\partial z}J_2-\frac{\partial M}{\partial y}J_3\,dA+ \iint_D\frac{\partial N}{\partial x}J_3-\frac{\partial N}{\partial z}J_1\,dA+ \iint_D\frac{\partial P}{\partial y}J_1-\frac{\partial P}{\partial x}J_2\,dA$

$\displaystyle \int_{\partial D} \left\langle M\frac{\partial x}{\partial u}+ N\frac{\partial y}{\partial u}+ P\frac{\partial z}{\partial u}, M\frac{\partial x}{\partial v}+ N\frac{\partial y}{\partial v}+ P\frac{\partial z}{\partial v}\right\rangle\cdot\langle du,dv\rangle$

This splits into three parts:

$\displaystyle \iint_D\frac{\partial M}{\partial z}J_2-\frac{\partial M}{\partial y}J_3\,dA= \int_{\partial D} M\frac{\partial x}{\partial u}du+M\frac{\partial x}{\partial v}dv$

$\displaystyle \iint_D\frac{\partial N}{\partial x}J_3-\frac{\partial N}{\partial z}J_1\,dA =\int_{\partial D} N\frac{\partial y}{\partial u}du+N\frac{\partial y}{\partial v}dv$

$\displaystyle \iint_D\frac{\partial P}{\partial y}J_1-\frac{\partial P}{\partial x}J_2\,dA= \int_{\partial D} P\frac{\partial z}{\partial u}du+P\frac{\partial z}{\partial v}dv$

By Green’s Theorem (applied in the ${(u,v)}$ -plane to the vector field ${(u,v)\mapsto \langle M\frac{\partial x}{\partial u},M\frac{\partial x}{\partial v} \rangle}$ ) gives:

$\displaystyle \int_{\partial D} M\frac{\partial x}{\partial u}du+M\frac{\partial x}{\partial v}dv=\iint_D \frac{\partial}{\partial u}\left(M\frac{\partial x}{\partial v}\right)- \frac{\partial}{\partial v}\left(M\frac{\partial x}{\partial u}\right)\,dA$

Now from the product rule

$\displaystyle \frac{\partial}{\partial u}\left(M\frac{\partial x}{\partial v}\right)=\frac{\partial M}{\partial u}\frac{\partial x}{\partial v}+M\frac{\partial^2 x}{\partial u\partial v}$

and

$\displaystyle \frac{\partial}{\partial v}\left(M\frac{\partial x}{\partial u}\right)=\frac{\partial M}{\partial v}\frac{\partial x}{\partial u}+M\frac{\partial^2 x}{\partial v\partial u}$

Since ${M\frac{\partial^2 x}{\partial u\partial v}=M\frac{\partial^2 x}{\partial v\partial u}}$ we have

$\displaystyle \frac{\partial}{\partial u}\left(M\frac{\partial x}{\partial v}\right)- \frac{\partial}{\partial v}\left(M\frac{\partial x}{\partial u}\right)=\frac{\partial M}{\partial u}\frac{\partial x}{\partial v}-\frac{\partial M}{\partial v}\frac{\partial x}{\partial u}$

Now the chain rule gives:

$\displaystyle \frac{\partial M}{\partial u}=\frac{\partial M}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial M}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial M}{\partial z}\frac{\partial z}{\partial u}$

$\displaystyle \frac{\partial M}{\partial v}=\frac{\partial M}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial M}{\partial y}\frac{\partial y}{\partial v}+\frac{\partial M}{\partial z}\frac{\partial z}{\partial v}$

A little computation gives

$\displaystyle \frac{\partial M}{\partial u}\frac{\partial x}{\partial v}-\frac{\partial M}{\partial v}\frac{\partial x}{\partial u}=-\frac{\partial M}{\partial y}\frac{\partial(x,y)}{\partial(u,v)}+\frac{\partial M}{\partial z}\frac{\partial(z,x)}{\partial(u,v)}$

So the above becomes

$\displaystyle \iint_D \frac{\partial}{\partial u}\left(M\frac{\partial x}{\partial v}\right)- \frac{\partial}{\partial v}\left(M\frac{\partial x}{\partial u}\right)\,dA=\iint_D \frac{\partial M}{\partial z}J_2-\frac{\partial M}{\partial y}J_3\,dA$

This is what we wanted for ${M}$ , the cases for ${N}$ and ${P}$ are similar. $\Box$

Just as we used Green’s theorem before we can use Stoke’s theorem to prove

Theorem 2 If ${F}$ is continuously differentiable in an open connected and simply connected region ${D}$ of ${{\mathbb R}^3}$ and ${\nabla\times F=0}$ , then ${F}$ is conservative.

Proof: (Hint.) Take ${A}$ and ${B}$ in ${D}$ and consider two paths ${P_1}$ and ${P_2}$ from ${A}$ to ${B}$ . Follow ${P_1}$ from ${A}$ to ${B}$ and ${P_2}$ back. This will result in some shared edges in which the fact that the orientation is reversed will cancel out and some polygons to which we can apply Stoke’s and get that the integral along the boundary is ${0}$ . thus we get ${\int_{P_1} F\cdot dr=\int_{P_2} F\cdot dr}$ . $\Box$

Example 1. Find ${\oint_C 2z\,dx+x\,dy+3y\,dz}$ where ${C}$ is the curve formed by the intersection of ${3x^2+y^2=z}$ and ${x+3y+4z=3}$ where the orientation is the outward normal.

Consider the planar surface ${S}$ with boundary ${C}$ and normal ${\vec n=\vec N/|\vec N|}$ where ${\vec N=\langle 1,3,4\rangle}$ . We have

$\displaystyle \oint_C 2z\,dx+x\,dy+3y= \iint_S \nabla \times \langle 2z,x,3y\rangle\cdot \vec n\,d\sigma$

$\displaystyle \det\left( \begin{array}{ccc} \hat i&\hat j&\hat k\\ \partial/\partial x &\partial/\partial y &\partial/\partial z \\ 2z&x&3y \end{array} \right)=\langle 3,2,1\rangle$

Viewing the plane as the level surface ${f(x,y,z)=3}$ we have

$\displaystyle \vec n\,d\sigma=\frac{\nabla f}{|f_z|}\,dA=\frac{\langle 1,9,16 \rangle}{4}\,dA$

So our integral is

$\displaystyle \iint_{S_{xy}} \langle 3,2,1\rangle\frac{\langle 1,9,16 \rangle}{4}\,dA$

Where ${S_{xy}}$ is the projection of ${S}$ in the ${xy}$ -plane. To find ${S_{xy}}$ we compute the intersection of the plane and ${z=3x^2+y^2}$ , this gives

$\displaystyle x+3y+4(3x^2+y^2)=3$

rearanging gives

$\displaystyle 3(x+1/24)^2+(y+3/8)^2=(1/4)[3+(1/24)^2+(3/8)^2]=905/1152$

This is an ellipse ${(x+1/24)^2/a^2+(y+3/8)^2/b^2=1}$ centered at ${(-1/24,-3/8)}$ with

$\displaystyle a=\sqrt{905/3456} \qquad b=\sqrt{905/1152}$

The area of this ellipse is ${ab\pi=(905/3456)(905/1152)\pi}$ So the integral is

$\displaystyle \frac{37}{4}\iint_{S_{xy}}\,dA=(37/4)(905/3456)(905/1152)\pi$

Problems:
14.7: 3, 5, 7, 9, 13, 19, 21, 25

Posted in Math 275, Spring 2009 | Leave a Comment »

Math 170 – May 1

May 2, 2009 by ketchers

We will skip 5.7 and spend the last week on application of integrals.

As far as problems go you need to work every integral in 5.5 and 5.6 – NO KIDDING – this will be a huge part of the final and if you can’t use the substitution rule you won’t do well. If a problem is too easy skip it, but please work the integrals that aren’t easy for you.

Posted in Math 170, Spring 2009 | Leave a Comment »

Math 275 – May 1

May 2, 2009 by ketchers

Here is a printable version of this post.

1. Surface Integrals

Just as for curves to develop a theory of integration on surfaces we need to parametrize the surface. A parametrization of a surface ${S}$ is just a map ${r:D\subseteq{\mathbb R}^2\rightarrow{\mathbb R}^3}$ that is continuous, 1-1, and onto ${S}$ . Often we write ${(x(u,v),y(u,v),z(u,v))}$ rather than ${r(u,v)}$ .

The level curves of ${r}$ are those curves in ${{\mathbb R}^3}$ that you get by fixing one of ${u}$ or ${v}$ and letting the other vary.

Example 1 – Torus:\label{torus} A parametrization of the torus centered at the origin with major radius ${R}$ and minor radius ${r}$ , assume ${R \ge r}$ , is given by: For ${0\le u,v \le 2\pi}$ set

$\displaystyle x(u,v)=(R+r\cos(v))\cos(u)\quad y(u,v)=(R+r\cos(v))\sin(u) \quad z(u,v)=r\sin(v)$

So ${s:[0,2\pi]\times[0,2\pi]\rightarrow S}$ is given by ${s(u,v)=\langle x(u,v),y(u,v),z(u,v) \rangle}$

Fixing ${v=v_0}$ the ${v_0}$ -level curve (so ${u}$ varies) is a circle parallel to the ${xy}$ -axis and in particular for ${v=0}$ this circle lies in the ${xy}$ -plane and has radius ${R}$ . Fixing ${u=u_0}$ the ${u_0}$ -level curve is a circle of radius ${r}$ centered on the the ${v=0}$ -level curve. In this way you can see the surface is a torus. Here is a picture with level curves drawn for ${v}$ and ${u}$ at ${0,\pi/6,\pi/3}$ , etc.

Example 2 – Mobius Strip: A parametrization of the Mobius strip centered at the origin with major radius ${R}$ and width ${2r}$ , assume ${R \ge r}$ , is given by: For ${0\le v \le 2\pi}$ and ${-r \le u \le r}$ set

$\displaystyle x(u,v)=(R+u\cos(v/2))\cos(u)\quad y(u,v)=(R+u\cos(v/2))\sin(u) \quad z(u,v)=u\sin(v/2)$

This parametrization describes taking a strip of paper of length ${2\pi R}$ and width ${2r}$ and twisting it so as to bring the short ends together with one twist. Here is a picture with level curves drawn for ${v}$ at ${0,\pi/6,\pi/3,\ldots,11\pi/6}$ and ${u}$ at ${-1,-3/4,-1/2,\ldots,3/4,1}$ .

A surface ${S}$ is smooth there is parametrization ${s:D \rightarrow S}$ so that

${D}$ has smooth boundary.
${s}$ has continuous first partials.
the vectors ${s_u=\langle x_u,y_u,z_u \rangle}$ and ${s_v=\langle x_v,y_v,z_v \rangle}$ are non-zero and not colinear.

The first condition implies that ${S}$ has a smooth boundary. The third condition is equivalent to ${s_u\times s_v\neq 0}$ . A surface ${S}$ is piecewise smooth if it is composed of a finite number of smooth pieces that intersect at most on their boundaries. The tangent plane to ${S}$ at ${s(u_0,v_0)}$ is given parametrically as

$\displaystyle s(u_0,v_0)+\alpha s_u(u_0,v_0)+\beta s_v(u_0,v_0) \text{ for }\alpha,\beta\in{\mathbb R}$

which, of course, is the same as the plane through ${s(u_0,v_0)}$ with normal ${s_u(u_0,v_0) \times s_v(u_0,v_0)}$ .

The transformation ${s:D \rightarrow S}$ transforms the rectangle ${[u_0,v_0]\times[u_0+\Delta u,v_0+\Delta v]}$ to a patch of the surface that has surface area ${\Delta\sigma}$ . ${\Delta\sigma}$ is approximated by the parallelogram given by the vectors ${s_u(u_0,v_0)\Delta u}$ and ${s_v(u_0,v_0)\Delta v}$ . So

$\displaystyle \Delta\sigma=|s_u\times s_v|\Delta u\Delta v$

This gives the differential

$\displaystyle d\sigma=|s_u\times s_v|\,du\,dv$

Often one variable is given in terms of the other two, for example ${z=z(x,y)}$ for ${(x,y)\in S_{xy}}$ where ${S_{xy}}$ is the projection of ${S}$ on the ${xy}$ -plane. In this case the parameterization is

$\displaystyle s:S_{xy}\rightarrow S\text{ given by }(x,y)\mapsto (x,y,z(x,y))$

In this case

$\displaystyle s_x=\langle 1,0,z_x\rangle\text{ and }s_y=\langle 0,1,z_y\rangle$

$\displaystyle s_x\times s_y=\langle -z_x,-z_y,1\rangle$

and

$\displaystyle d\sigma=\sqrt{z_x^2+z_y^2+1}\,dx\,dy= \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+ \left(\frac{\partial z}{\partial y}\right)^2+1}\,dx\,dy$

This might be remembered as follows, recall

$\displaystyle \nabla=\left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle$

$\displaystyle |\nabla|=\sqrt{\left(\frac{\partial}{\partial x}\right)^2+ \left(\frac{\partial}{\partial y}\right)^2+ \left(\frac{\partial}{\partial z}\right)^2}$

and

$\displaystyle d\sigma=|\nabla|\,dx\,dy\,dz$

This is really a bunch of formalism, but it works out nicely.

Given a real valued function ${f:S\rightarrow{\mathbb R}}$ we define the surface integral of ${f}$ on ${S}$ by fixing a parametrization ${r:D \rightarrow S}$ then partitioning ${D}$ into little rectangles ${\Delta D}$ which get transformed into little patches of the surface ${\Delta S}$ whose surface area is ${\Delta \sigma}$ . Thus we have

$\displaystyle \iint_S f \, d\sigma$

In particular the surface area of ${S}$ is

$\displaystyle \iint_S \, d\sigma$

If ${\delta(x,y,z)}$ is a density function (for a thin sheet ${S}$ ), then the mass is ${M=\iint_S \delta\,d\sigma}$ , similarly for first moments, moments of inertia, etc.

Example 3. Find the surface area of the \hyperlink{torus}{torus}. We have

$\displaystyle s_u=(R+r\cos(v))\langle -\sin(u),\cos(u) ,0 \rangle$

$\displaystyle s_v=-r\langle \sin(v)\cos(u),\sin(v)\sin(u),-\cos(v)\rangle$

$\displaystyle s_u\times s_v=(-r)(R+r\cos(v))\det\left( \begin{array}{ccc} \hat i&\hat j&\hat k\\ -\sin(u) & \cos(u) &0\\ \sin(v)\cos(u) & \sin(v)\sin(u)&-\cos(v) \end{array} \right)$

$\displaystyle =(-r)(R+r\cos(v))\langle -\cos(u)\cos(v),-\sin(u)\cos(v),-\sin(v)$

$\displaystyle |s_u\times s_v|=Rr + r\cos(v)$

and

$\displaystyle \iint_S d\sigma=\int_0^{2\pi}\int_0^{2\pi}Rr + r\cos(v) \, dv\,du=(2\pi R)(2\pi r)$

This can be seen to be correct geometrically by cutting it and stretching it out into a cylinder of height ${2\pi R}$ with radius ${r}$ .

If a surface is given as a level curve ${f(x,y,z)=C}$ with continuous first partials and ${\nabla f\neq 0}$ , then at each point ${P}$ on ${S}$ we can find an open neighborhood of ${P}$ so that one variable is given in terms of the other two – hence a local parametrization. This follows from the implicit function theorem which in this case can be stated as

Theorem 1 Assume ${f}$ has continuous partials in a nbhd of ${P}$ and ${f_z(P) \neq 0}$ , then there is a nbhd ${U}$ of ${P}$ so that ${z=h(x,y)}$ for some ${h:U\rightarrow {\mathbb R}}$ that is differentiable on ${U}$ . (Similarly we can solve for ${x}$ or ${y}$ if the appropriate partial is non-zero.)

Since ${h}$ is differentiable we can apply the chain rule using ${s(x,y)=(x,y,h(x,y))}$ :

$\displaystyle g(x,y)=f(x,y,h(x,y))=C$

for ${x,y}$ in ${U}$ . So

$\displaystyle g_x=f_x+f_z\frac{\partial h}{\partial x}=0$

$\displaystyle g_y=f_y+f_z\frac{\partial h}{\partial y}=0$

$\displaystyle \frac{\partial h}{\partial x}=-\frac{f_x}{f_z}\quad \frac{\partial h}{\partial y}=-\frac{f_y}{f_z}$

This means

$\displaystyle s_x=(1,0,-f_x/f_z)\quad s_y=(0,1,-f_y/f_z)$

and

$\displaystyle s_x\times s_y=\left\langle \frac{f_x}{f_z},\frac{f_y}{f_z},1\right\rangle =\frac{\langle f_x,f_y,f_z\rangle}{f_z} =\frac{\nabla f}{f_z} = \frac{\nabla f}{\nabla f\cdot \hat k}$

$\displaystyle d\sigma=\frac{|\nabla f|}{|f_z|}\,dx\,dy=\frac{|\nabla f|}{|\nabla f\cdot \hat k|}\,dx\,dy$

and hence

$\displaystyle \iint_S g(x,y,z) \,d\sigma =\iint_D g(x,y,z)\frac{|\nabla f|}{|f_z|}\,dA$

where ${D}$ is the “shadow” of ${S}$ in the ${xy}$ -plane. (This is assuming ${f_z\neq 0}$ , the same sort of expression works for projections in the other coordinate planes assuming the appropriate partial is non-zero.)

Example 4. Suppose the density of a hemispherical dome of radius ${r}$ is ${4z}$ , find the mass. The surface ${S}$ is given by ${f(x,y,z)=x^2+y^2+z^2=16}$ . The projection of ${S}$ on the ${xy}$ -plane is the disk ${D}$ of radius ${4}$ , ${x^2+y^2\le 4}$ . Here ${\nabla f=\langle 2x,2y,2z\rangle}$ so ${|\nabla f|=\sqrt{4x^2+4y^2+4z^2}=2\sqrt{x^2+y^2+z^2}=(2)(4)=8}$ and ${|f_z|=2|z|=2z}$ since ${z\ge 0}$ so

$\displaystyle \iint_S 4z\, d\sigma=\iint_D 4z \frac{|\nabla f|}{|f_z|}\,dA= \iint_D 4z(8/z)\,dA=32\int_0^{2\pi}\int_0^4r\,dr\,d\theta$

2. Flux through a surface

A surface is orientable if it is possible to continuously (in the variables ${x,y,z}$ ) assign a normal unit vector ${\vec n}$ to each point on the surface.

If ${S}$ is smooth as witnessed by ${s:D\rightarrow S}$ and ${S}$ is orientable, then

$\displaystyle \vec n=\pm\frac{s_u\times s_v}{|s_u\times s_v|}$

This need not be the case if ${S}$ is non-orientable as can be seen by the Mobius strip, which in non-orientable. The point is that ${\frac{s_u\times s_v}{|s_u\times s_v|}}$ is continuous as a function on ${[0,2\pi]\times[0,2\pi]}$ (in the ${u,v}$ -plane), however, the induced assignment of normal vectors on the Mobius strip in ${{\mathbb R}^3}$ as a function of ${(x,y,z)}$ is not continuous as is seen below.

If ${S}$ is given as the level surface to a continuously differentiable ${f}$ with ${\nabla f\neq 0}$ , then

$\displaystyle \vec n=\pm\frac{\nabla f}{|\nabla f|}$

If ${S}$ is orientable with orientation ${\vec n}$ and ${F:{\mathbb R}^3\rightarrow{\mathbb R}^3}$ is a continuous vector field defined on a region including ${S}$ , then

$\displaystyle \vec n\,d\sigma=\pm\frac{s_u\times s_v}{|s_u\times s_v|}|s_u\times s_v|\,du\,dv=\pm(s_u\times s_v)\,du\,dv$

so the flux through ${S}$ is

$\displaystyle \iint_S F\cdot \vec n\,d\sigma=\pm\iint_D F \cdot (s_u\times s_v)\,du\,dv$

If ${z=z(x,y)}$ on ${S_{xy}}$ , then

$\displaystyle \text{flux}=\pm\iint_{S_{xy}} F \cdot\langle -z_x,-z_y,1\rangle\,dx\,dy$

Similarly for ${x=x(y,z)}$ on ${S_{yz}}$ and ${y=y(x,z)}$ on ${S_{x,z}}$ .

If ${S}$ is given as a level surface ${f(x,y,z)=c}$ and ${f_z\neq 0}$ in ${S_{xy}}$ , then

$\displaystyle d\sigma=\pm\frac{\nabla f}{|\nabla f|} \frac{|\nabla f|}{|f_z|}|\,dx\,dy = \pm\frac{\nabla f}{|f_z|}\,dx\,dy$

Similarly with ${x}$ or ${y}$ replacing ${z}$ . So

$\displaystyle \text{flux}=\pm\iint_{S_{xy}} F\cdot \frac{\nabla f}{|f_z|}\,dx\,dy$

Example 5. Find the flux of the field ${F(x,y,x)=\langle x^2,y^2,z\rangle}$ through the hemisphere of radius ${2}$ lying above the ${xy}$ -plane. The surface ${S}$ is given by ${f(x,y,z)=x^2+y^2+z^2=4}$ . Here the normal ${\vec N=\frac{\nabla f}{|\nabla f|}}$ . ${\nabla f=\langle 2x,2y,2z\rangle}$ so ${|\nabla f|=2\sqrt{x^2+y^2+z^2}=(2)(2)=4}$ and so

$\displaystyle \text{flux}=\iint F\cdot \vec n\,d\sigma=\frac{1}{4}\iint_D \langle x^2,y^2,z\rangle\cdot\langle 2x,2y,2z\rangle \, dA$

$\displaystyle =\frac{1}{4}\iint_D \left[ 2x^3+2y^3+2z^2 \right]\,dA =\frac{1}{2}\iint_D \left[ x^3+y^3+(4-x^2-y^2) \right]\,dA$

Here ${D}$ is the disk of radius ${2}$ on the ${xy}$ -plane centered at the origin, ${x^2+y^2\le 2}$ . So

$\displaystyle \text{flux}=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \left[ x^3+y^3+(4-x^2-y^2) \right]\,dy\,dx$

Example 6. Find flux of the gradient field of ${f(x,y,z)=xyz}$ through the surface ${S}$ given by ${z=4-(x^2-y^2)}$ over the triangular region ${D}$ in the ${xy}$ -plane with corners ${(0,0)}$ , ${(1,3)}$ , ${(2,-5)}$ .