Question

[CALCULUS 3: SURFACE INTEGRALS & APPLICATIONS]

Please see the question posted below.

Answer 1

Use Stoke's Theorem to evaluate \[\int\limits_{C}^{ }F \times dr\]where \[F = <4x + 7y, y + 5z, 9z + 2x>\]and C is the triangle with vertices (2, 0, 0), (0, 6, 0) and (0, 0, 12) oriented so that the vertices are transversed in the specified order.

Answer 2

Ok, show me your best attempt. I suggest drawing out the region to help you out.

Answer 3

use `\cdot` for dot product

Answer 4

I really don't know where to begin with this. My main problem in Calc 3 is visualizing the region that these integrals are projected on. Right now, I'm just breaking up the F-vector into components of P, Q and R and performing a curl on F to obtain the function to plug into my double integral. But I'm really not sure where to go from the curl F.

Answer 5

To find projections, think of shining a flashlight from the center of the function on to each of the 3 planes (i.e xy-plane, xz-plane, and yz-planes) Just a hint.

Answer 6

That's the problem, you're trying to visualize it. Don't think so hard, just draw it! Then you have the picture to work with instead of trying to guess in your mind.

Answer 7

So I'm able to draw out the triangle obviously, and I think I drew out the vector field properly. But what from there? Should I make a triple integral for the dimensions of x y and z from the triangle? I don't think I have to reparametrize this, but I'm not too sure.

Answer 8

Stokes theorem translates a line integral(distance problem) to a surface integral (area problem)

Answer 9

you need to choose a surface whose boundary is the closed path along which you're trying to calculate the work

Answer 10

you can choose any surface whose boundary is the triangle

Answer 11

\[\oint_C \vec{F}\cdot \mathrm{d}\vec{r} = \iint_S (\nabla \times \vec{F})
\cdot \vec{n} dS\]

Answer 12

choose your favorite surface and calculate the flux through it

that equals the work done along the triangle by stokes thm

Answer 13

Since this is a plane, it seems easiest to pick this as the surface, although it is just as valid to just do the triangles that are on the axes, but it seems like you'd be doing essentially 3 times the effort. To get the normal vector of the plane you just need to take the cross product of two vectors, like the one that goes from the point (2,0,0) to (0,6,0) and (2,0,0) to (0,0,12).

It should make sense from your picture. Just draw it on here so I can help you.

Answer 14

Definitely easier to work with your domain first (since it's transversed) than work on finding the components (P, Q, R) of your function. :|

Answer 15

I'm sorry, guys. I'm just not getting this. I've solved for the curl F vector to be (-5, -2, -7). I've solved for a normal vector of the triangle without transversing it on any surface to be 72x + 22y + 10z - 144 = 0. I don't know how to transverse this on the surface S given by F and I don't know how to calculate the region to integrate the function.

Answer 16

curl is right but the equation of plane looks wrong

Answer 17

(2, 0, 0), (0, 6, 0) and (0, 0, 12)

you're given intercepts of plane so finding the equation of plane is trivial :
\[\dfrac{x}{2} + \dfrac{y}{6} + \dfrac{z}{12} = 1\]

which is same as
\[6x+2y+z = 12\]

Answer 18

`oriented so that the vertices are transversed in the specified order. `

this just tells you that you need to choose the orientation of plane with normal vector sticking up/away from the origin (use right hand rule)

Answer 19

the answer changes by a sign if you choose a wrong orientation. so its okay, don't let this orientation stuff keep you from finishing the problem. just keep going... you will get how to choose correct orientation by practicing few more problems

Answer 20

\[\begin {align} \oint_C \vec{F}\cdot \mathrm{d}\vec{r} &= \iint_S (\nabla \times \vec{F})\cdot \vec{n} \mathrm{d}S \\~\\

&= \iint_S \langle -5,-2,-7\rangle \cdot \langle 6,2,1 \rangle \mathrm{d}x\mathrm{d}y \\~\\

\end{align}\]


setup the bounds for the "shadow" of surface bounded by triangle in xy plane and evaluate the integral

Answer 21

Does it matter that it is in the xy-plane? Could I set the bounds relative to dydz or dxdz or any variant of the three if I make the triangle's bounds correct?

Answer 22

it doesn't matter if you're careful :
\[\vec{n}\mathcal{d}S = \dfrac{\vec{N}}{\vec{N}\cdot \hat{k}}\mathcal{d}x\mathcal{d}y = \dfrac{\vec{N}}{\vec{N}\cdot \hat{i}}\mathcal{d}y\mathcal{d}z= \dfrac{\vec{N}}{\vec{N}\cdot \hat{j}}\mathcal{d}z\mathcal{d}x\]

Answer 23

just use the appropriate conversion formula

Answer 24

actually for this particular problem, you don't need to use integrals if you are good with geometry

Answer 25

I got my final answer to be -246, which is correct. Thank you guys so so much, I can't begin to explain how much this helps. I've been just looking at the whole picture completely off. Some more practice should be good for me. Thank you very very much.

Answer 26

Nice :) notice that whenever the integrand is constant, life becomes simple. you can simply multply the area of shadow by the integrand :

\[\begin {align} \oint_C \vec{F}\cdot \mathrm{d}\vec{r} &= \iint_S (\nabla \times \vec{F})\cdot \vec{n} \mathrm{d}S \\~\\

&= \iint_S \langle -5,-2,-7\rangle \cdot \langle 6,2,1 \rangle \mathrm{d}x\mathrm{d}y \\~\\

&= -41\iint_S \mathrm{d}x\mathrm{d}y \\~\\
&= -41\times (\text{Area of shadow in xy plane}) \\~\\
&= -41\times \left(\dfrac{1}{2}\times 2\times 6\right) \\~\\
\end{align}\]