Math 214 - Multivariable Calculus - Prep

Task 45.1

We've discussed circulation density and connected it to the curl of a vector field. Let's construct a corresponding analog for flux.

Given a point $(x,y,z)$, we calculate the flux density of $\vec F$ at $(x,y,z)$ as follows.

We construct a sphere $S_a$ of radius $a$ centered at $(x,y,z)$.
We compute the outward flux of $\vec F$ across the surface of the sphere, so $\iint_{S_a} \vec F\cdot\hat n dS$.
We divide by the volume $V_a$ inside $S_a$, to obtain a flux per volume.
The flux density of $\vec F$ at $(x,y,z)$ is the limit $$\lim_{a\to 0}\frac{1}{V_a}\iint_{S_a}\vec F\cdot\hat n dS = \vec \nabla \cdot \vec F.$$

The divergence of a vector field provides the flux density.

We are now ready to state the divergence theorem.

Divergence Theorem

Given a solid domain $D$ with piecewise smooth boundary $\partial D$, where each portion of the boundary is a surface oriented pointing outwards away from $D$, then the integral of the divergence of $\vec F=(M,N,P)$ along $D$ is equal to the outward flux of $\vec F$ across the boundary $\partial D$, which we can write as $$\iiint_D \vec \nabla \cdot \vec F dV = \iint_{\partial D}\vec F\cdot\hat n dS.$$

Note that $\partial D$ might consist of multiple surfaces. Sum $\iint_{S}\vec F\cdot\hat n dS$ for each surface $S$ that is part of the boundary to compute $\iint_{\partial D}\vec F\cdot\hat n dS$.
For every surface $S$ that is part of the boundary, remember to make sure $\hat n$ is the outward pointing unit vector, pointing away from the solid domain $D$.

Let's verify this theorem in some examples.

Consider the vector field $\vec F = (x^2, 2 y, x + y + z)$ and sphere of radius 5. We can parameterize the surface $S$ using $\vec r(u,v) = (5 \cos(u)\sin(v),5\sin(u)\sin(v),5\cos(v))$ for $0\leq u\leq 2\pi$ and $0\leq v\leq \pi$. The solid domain $D$ satisfies $\rho\leq 5$ (using spherical coordinates - you'll need appropriate bounds for $\theta$ and $\phi$). Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $500\pi$ for both. Pay attention to the direction in which $\vec r_u\times \vec r_v$ points.)
Consider the vector field $\vec F = (5x,x-z,y+z+1)$ and solid domain $D$ in space that lies below the paraboloid $z=1-r^2$ (using cylindrical coordinates) and above the $xy$-plane. The boundary of $D$ consists of two surfaces. The top surface is parametrized by $\vec r(u,v) = (u \cos v,u \sin v,1-u^2)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. The bottom surface is the plane $z=0$, and for this particular problem we can parameterize it using $\vec r(u,v) = (u\cos v,u\sin v,0)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $3\pi$ for both. The flux integral across the top surface is $4\pi$, and across the bottom surface is $-\pi$.)

Task 45.2

Stokes's theorem can be applied when the boundary of a surface consists of disjoint curves. Remember to check if the parametrization associated with each curve is compatible with the orientation on the surface. When not compatible, a simple negative sign can quickly fix any issues.

Let $\vec F = (2x-3y,x^2+2z,4xy)$.

Consider the surface $S$ parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq 2\pi$, oriented by $\hat n$ pointing away from the $y$-axis. The boundary $\partial S$ of $S$ is the union of the two disconnected curves $C_1$ with parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq 2\pi$ and $C_2$ with parametrization $\vec r_2(t) = (3\cos t, 3, 3\sin t)$ for $0\leq t\leq 2\pi$. Provided the curves are oriented compatibly with $\hat n$, Stokes's theorem states that $$\ds \int_{\partial S} Mdx+Ndy+Pdz = \iint_S \vec \nabla \times \vec F \cdot \hat n dS.$$
1. Draw the surface and curves.
2. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$. (I got $104\pi$.)
3. Compute $\ds \int_{C_1} Mdx+Ndy+Pdz$ and $\ds \int_{C_2} Mdx+Ndy+Pdz$, using the parametrizations provided.
4. Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$.
Now instead let $S$ be half of the surface above, parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq \pi$. The boundary of $S$ now consists of 4 connected curves, two arcs and two straight lines. One arc has parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq \pi$. One of the straight lines has parametrization $\vec r_2(t) = (t, t, 0)$ for $1\leq t\leq 3$.
1. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$ (change the bounds of what you did in the previous part).
2. Give parametrizations for the other two curves that make up $\partial S$.
3. Compute $\ds \int_{C} Mdx+Ndy+Pdz$ for each of the 4 curves in $\partial S$. (You'll end up with $6+2\pi$, $54+54\pi$, $14/3$, and $86/3$, all of which could be negative instead of positive, based on your parametrization.)
4. Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$.

Task 45.3

At this point, we've seen all the topics for the semester, and just need to make sure we practice them all. If you haven't yet, please create blocks of Mathematica code that will help you quickly compute work, surface integrals, and flux integrals, as well as quickly see $\vec r_u\times \vec r_v$ so you can make decisions about in which direction this normal vector points. I'll add more problems for later days. If you want more to work on before then, head to OpenStax and tackle any of the problems in 6.6, 6.7, or 6.8.

Task 45.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.

Big View
Prep Day 45 Task 45.1 We've discussed circulation density and connected it to the curl of a vector field. Let's construct a corresponding analog for flux. Given a point $(x,y,z)$, we calculate the flux density of $\vec F$ at $(x,y,z)$ as follows. We construct a sphere $S_a$ of radius $a$ centered at $(x,y,z)$. We compute the outward flux of $\vec F$ across the surface of the sphere, so $\iint_{S_a} \vec F\cdot\hat n dS$. We divide by the volume $V_a$ inside $S_a$, to obtain a flux per volume. The flux density of $\vec F$ at $(x,y,z)$ is the limit $$\lim_{a\to 0}\frac{1}{V_a}\iint_{S_a}\vec F\cdot\hat n dS = \vec \nabla \cdot \vec F.$$ The divergence of a vector field provides the flux density. We are now ready to state the divergence theorem. Divergence Theorem Given a solid domain $D$ with piecewise smooth boundary $\partial D$, where each portion of the boundary is a surface oriented pointing outwards away from $D$, then the integral of the divergence of $\vec F=(M,N,P)$ along $D$ is equal to the outward flux of $\vec F$ across the boundary $\partial D$, which we can write as $$\iiint_D \vec \nabla \cdot \vec F dV = \iint_{\partial D}\vec F\cdot\hat n dS.$$ Note that $\partial D$ might consist of multiple surfaces. Sum $\iint_{S}\vec F\cdot\hat n dS$ for each surface $S$ that is part of the boundary to compute $\iint_{\partial D}\vec F\cdot\hat n dS$. For every surface $S$ that is part of the boundary, remember to make sure $\hat n$ is the outward pointing unit vector, pointing away from the solid domain $D$. Let's verify this theorem in some examples. Consider the vector field $\vec F = (x^2, 2 y, x + y + z)$ and sphere of radius 5. We can parameterize the surface $S$ using $\vec r(u,v) = (5 \cos(u)\sin(v),5\sin(u)\sin(v),5\cos(v))$ for $0\leq u\leq 2\pi$ and $0\leq v\leq \pi$. The solid domain $D$ satisfies $\rho\leq 5$ (using spherical coordinates - you'll need appropriate bounds for $\theta$ and $\phi$). Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $500\pi$ for both. Pay attention to the direction in which $\vec r_u\times \vec r_v$ points.) Consider the vector field $\vec F = (5x,x-z,y+z+1)$ and solid domain $D$ in space that lies below the paraboloid $z=1-r^2$ (using cylindrical coordinates) and above the $xy$-plane. The boundary of $D$ consists of two surfaces. The top surface is parametrized by $\vec r(u,v) = (u \cos v,u \sin v,1-u^2)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. The bottom surface is the plane $z=0$, and for this particular problem we can parameterize it using $\vec r(u,v) = (u\cos v,u\sin v,0)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $3\pi$ for both. The flux integral across the top surface is $4\pi$, and across the bottom surface is $-\pi$.) Task 45.2 Stokes's theorem can be applied when the boundary of a surface consists of disjoint curves. Remember to check if the parametrization associated with each curve is compatible with the orientation on the surface. When not compatible, a simple negative sign can quickly fix any issues. Let $\vec F = (2x-3y,x^2+2z,4xy)$. Consider the surface $S$ parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq 2\pi$, oriented by $\hat n$ pointing away from the $y$-axis. The boundary $\partial S$ of $S$ is the union of the two disconnected curves $C_1$ with parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq 2\pi$ and $C_2$ with parametrization $\vec r_2(t) = (3\cos t, 3, 3\sin t)$ for $0\leq t\leq 2\pi$. Provided the curves are oriented compatibly with $\hat n$, Stokes's theorem states that $$\ds \int_{\partial S} Mdx+Ndy+Pdz = \iint_S \vec \nabla \times \vec F \cdot \hat n dS.$$ Draw the surface and curves. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$. (I got $104\pi$.) Compute $\ds \int_{C_1} Mdx+Ndy+Pdz$ and $\ds \int_{C_2} Mdx+Ndy+Pdz$, using the parametrizations provided. Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$. Now instead let $S$ be half of the surface above, parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq \pi$. The boundary of $S$ now consists of 4 connected curves, two arcs and two straight lines. One arc has parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq \pi$. One of the straight lines has parametrization $\vec r_2(t) = (t, t, 0)$ for $1\leq t\leq 3$. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$ (change the bounds of what you did in the previous part). Give parametrizations for the other two curves that make up $\partial S$. Compute $\ds \int_{C} Mdx+Ndy+Pdz$ for each of the 4 curves in $\partial S$. (You'll end up with $6+2\pi$, $54+54\pi$, $14/3$, and $86/3$, all of which could be negative instead of positive, based on your parametrization.) Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$. Task 45.3 At this point, we've seen all the topics for the semester, and just need to make sure we practice them all. If you haven't yet, please create blocks of Mathematica code that will help you quickly compute work, surface integrals, and flux integrals, as well as quickly see $\vec r_u\times \vec r_v$ so you can make decisions about in which direction this normal vector points. I'll add more problems for later days. If you want more to work on before then, head to OpenStax and tackle any of the problems in 6.6, 6.7, or 6.8. Task 45.4 Pick some problems related to the topics we are discussing from the Text Book Practice page. Edit Page History Source Attach File Backlinks List Group Page last modified on November 30, 2022, at 02:38 PM
skin config pmwiki-2.2.142

Day 45

Task 45.1

Task 45.2

Task 45.3

Task 45.4