- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is day 4 of Unit 4.
Brain Gains (Rapid Recall, Jivin' Generation)
- Find the work done by the vector field $\vec F(x,y,z) = (2x+3z, 2yz, 3x+y^2)$ on an object that moves from $(0,0,0)$ to $(1,2,3)$.
Solution
A potential for the vector field is $f(x,y,z) = x^2+3xz+y^2z$. As such, the work done by $\vec F$ is the difference in potential, which gives $$W = f(1,2,3)-f(0,0,0) = (1+3+12)-(0) = 16.$$
Note that no path was given for this problem. Without a given path, had the vector field not had a potential then we could not compute the work done (as work depends on the path for vector fields without a potential).
When a vector field has a potential, work is path independent. When work done by a vector field is independent of path, we call the vector field conservative. As such, every vector field that has a potential is a conservative vector field.
- For a vector field $\vec F(x,y)=(M,N)$ that is continuously differentiable everywhere, Green's theorem states $$\oint_C Mdx+Ndy = \iint_R N_x-M_ydA,$$ where $R$ is region inside a simple (does not cross itself) closed (starts and ends at the same place) piecewise smooth (can be parametrized with finitely many differentiable segments) curve $C$ that traverses around the boundary of $R$ in a counterclockwise fashion. The circle on the integral states that the curve is closed.
- Let $\vec F = (3x+4y,6x+7y)$ and let $C$ be the curve which starts at $(1,-3)$, travels right to $(5,-3)$, up to $(5,7)$, left to $(1,7)$, and then back down to $(1,-3)$. Find the work done by $\vec F$ along $C$ using Green's Theorem.
Solution
Note that $N_x-M_y=6-4=2$. By Green's theorem, the requested integral is the same as $\iint_R2dA=2A$, twice the area of the rectangle $R$ inside $C$. The width of this rectangle is 4, and the height is 10, so the answer is $W=2A = 2(40)=80$.
- Draw the surface $S$ with parametrization $\vec r(u,v) = (u, u\cos v, u\sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq \pi$.
- Set up the surface integral that would give the surface area of $S$.
- Set up the surface integral formula that would give the $z$-coordinate of the centroid of the surface $S$.
- Set up the surface integral that gives the flux $\Phi = \iint_S \vec F\cdot \hat n dS$ of the vector field $\vec F(x,y,z) = (2x+y,-yz^2, 3z)$ across the surface $S$ in the direction that points away from the $x$-axis.
Solutions
The solutions are all coded using the Mathematica code below. Note that the vector $n$ has a positive $x$-component, which means that the vector points inward towards the $y$-axis. The negative sign in the code below is needed to correctly get the flux.
r[u_, v_] := {u, u Cos[v], u Sin[v]}
uB = {u, 1, 3}
vB = {v, 0, Pi}
(*Plot the surface*)
surfacePlot = ParametricPlot3D[r[u, v], uB, vB]
(*Compute the normal vector*)
n[u_, v_] := Cross[D[r[u, v], u], D[r[u, v], v]] // Simplify
n[u, v]
Norm[n[u, v]] // ComplexExpand // FullSimplify
(*Find the surface area.*)
surfaceArea = Integrate[Norm[n[u, v]], uB, vB]
(*Compute zbar*)
zBar = Integrate[u Sin[v] Norm[n[u, v]], uB, vB]/
Integrate[Norm[n[u, v]], uB, vB]
zBar = Integrate[r[u, v][[3]] Norm[n[u, v]], uB, vB]/
Integrate[Norm[n[u, v]], uB, vB]
(*Find the flux in three ways.*)
(*First do all the substition by hand.*)
Flux = Integrate[{2 u + u Cos[v], -u Cos[v] (u Sin[v])^2, 3 u Sin[v]} . -n[u, v], uB, vB]
(*Use ReplaceAll to preform the substution.*)
Flux = Integrate[ReplaceAll[{2 x + y, -y z^2, 3 z} . -n[u, v], {x -> u, y -> u Cos[v], z -> u Sin[v]}], uB, vB]
(*Use function compostion @@ to do all the substition.*)
F[x_, y_, z_] := {2 x + y, -y z^2, 3 z}
F @@ r[u, v]
F @@ r[u, v] . -n[u, v]
Integrate[F @@ r[u, v] . -n[u, v], uB, vB]
(*Show the surface along with the vector field to visualize the flux.*)
Show[surfacePlot,
VectorPlot3D[F[x, y, z], {x, 1, 3}, {y, -3, 3}, {z, 0, 3},
VectorScaling -> Automatic, VectorSizes -> Automatic,
VectorMarkers -> Placed["Arrow3D", "Start"]]
]
(*Use SliceVectorPlot3D to show just the vectors on the surface.*)
surface = ParametricRegion[r[u, v], Evaluate[{uB, vB}]];
SliceVectorPlot3D[
F[x, y, z], surface, {x, 1, 3}, {y, -3, 3}, {z, 0, 3},
VectorScaling -> Automatic, VectorSizes -> Automatic,
VectorMarkers -> Placed["Arrow3D", "Start"], BoxRatios -> {2, 6, 3}]
Group Problems
- Compute the work done by $\vec F = (-3y,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t).$
- Compute this work using $ \oint_{C} M dx+Ndy$. [Check $150\pi$.]
- Compute this work using $\iint_R N_x-M_y dA$. (You can set up the integral in rectangular, or polar, or just use facts about area.)
- Compute the work done by $\vec F = (2x-y,2x+4y)$ to move an object counterclockwise once along the triangle with corners $(0,0)$, $(2,0)$, and $(0,3)$.
- Set up the single double integral $\iint_R N_x-M_y dA$.
- Compute the integral (use use facts about area).
- Consider the surface $S$ parametrized by $\vec r(u,v) = (u\cos v, u\sin v, u^2)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$.
- Draw the surface.
- Compute $dS = \left|\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial v}\right|dudv$.
- Set up an integral formula to compute the surface area of $S$.
- Set up an integral formula to compute $\bar z$ for this surface.
- We would like an orientation $\hat n$ for the surface that points away from the $z$-axis. Does $ \dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial v}$ point towards the $z$-axis, or away from the $z$-axis?
- Set up the surface integral that gives the flux of $\vec F = (3yz,-2x+y, z-2x)$ across the surface $S$ in the direction of $\hat n$. Then use software to compute the integral.
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