Question

Let S be the portion of the paraboloid z = 4 -x^2 -y^2 that lies above z= 0. Let F be the vector field F = (z-y)i + (x+z)j - (e^xyz(cosy))k. Using Stoke's Theorem, find the surface integral.

Answer 1

So the hope is that \(\nabla \times \mathbf F\) is a very simple function, if we're using Stokes.

Answer 2

Agreed

Answer 3

Can you calculate it?

Answer 4

Yes. Let me re-calculate to make sure my answer isn't fudged.

Answer 5

\[(xze ^{xyz}cosy -e^{xyz}siny - 1)i -(e^{xyz}cosy - 1)j + 2k\]

Answer 6

What a mess.

Answer 7

Yes

Answer 8

okay, do you know how to parametrize the parabaloid then?

Answer 9

Maybe...

Answer 10

subtract z in the equation, and take the gradient of the expression.

Answer 11

-2x-2y, or -2x-2y-1? I dont quite understand what you meant by subtract z or why it's done that way.

Answer 12

the gradient is a vector

Answer 13

When you have a level set \[
g(x,y,z) = C
\]Then the gradient will give you the normal vector to that level set.

Answer 14

So you're saying that \[g(x,y,z) = C = 4 - x^2 -y^2 - z\]
\[gradient = (-2x)i - (2y)j - 1k \]
?
Sorry, this is again unfamiliar to me.

Answer 15

yes

Answer 16

Ok

Answer 17

you could have even said \[
g(x,y,z)=-4 = -x^2-y^2-z
\]It doesn't matter since \(C\) goes away when differentiated.

Answer 18

Excellent

Answer 19

Seems like a tough integral...

Answer 20

Would you use \[\int\limits_{}^{}\int\limits_{S}^{} (curlF) ndS = \int\limits_{C}^{} Fdr \]

Answer 21

Where \[Fdr = (z-y)dx + (x+z)dy + (e^{xyz}cosy)dz\]

Answer 22

In doing so, you are no longer using stokes.

Answer 23

Gotcha

Answer 24

\[dS = \sqrt(4x^2 + 4y^2 +1) ??\]

Answer 25

No, okay, so basically \(\mathbf n\) is the unit normal vector, and \(dS\) is the magnitude of the the normal vector \(d\mathbf S\), which you just calculated.

Answer 26

Well actually, I'm being a bit sloppy here.

Answer 27

Anyway \(d\mathbf S = \langle -2x,-2y,-1\rangle dxdy\)

Answer 28

And \[
\mathbf n~ dS = d\mathbf S
\]

Answer 29

Ok

Answer 30

So do we leave the z's as constants the whole time? or set all z's to 0?

Answer 31

Okay, so basically, the \(z\) gets replaced with \(z(x,y)\) when we do this sort of parametrization.

Answer 32

Since \(z=f(x,y)\).

Answer 33

So, for every z, we replace with 4 -x^2 - y^2 ?

Answer 34

yes

Answer 35

Are you sure you wrote the problem down correctly?

Answer 36

Very sure.

Answer 37

Our only hope is that cylindrical coords make things easier.

Answer 38

\[(x(4-x^2-y^2)e^{((4-y^2-x^2)xy)}cosy- e^{(4-x^2-y^2)yx}siny -1 )i -(y(4-x^2-y^2)e^{(4-x^2-y^2)yx}cosy-1)j +2k\]

Answer 39

Doesnt even fit. -_-

Answer 40

How should I go about using cyl coords for this? The only I've been to taught to use this is to convert it to a line integral since Stokes' Theorem allows for you to choose whichever is more convenient, which seems like the line integral would be for this particular instance

Answer 41

I should have said polar coords. \[
r^2 = x^2+y^2\\
\theta = \tan\frac yx\\
x = r\cos\theta\\
y = r\sin\theta \\
dx~dy = r~dr~d\theta
\]

Answer 42

Bleh. Either conversion is just... tedious. But given that....\[(\cos(\theta)(4r-r^3) e^{(4r^2-r^4)\cos(\theta)\sin(\theta)}\cos(rsin(\theta)) - e^{(4r^2-r^4)\cos(\theta)\sin(\theta)}\sin(rsin(\theta) - 1)i -(\sin(\theta)(4r-r^3)e^{(4r^2-r^4)\cos(\theta)\sin(\theta)}-1)j +2k\]

Answer 43

There is something wrong with this problem.

Answer 44

lol

Answer 45

I swear that F(x,y,z) = (z-y)i +(x+z)j - (e^xyz (cosy))k.
S is the portion of the paraboloid z = 4 - x^2 - y^2 that lies above z=0.

Answer 46

Given Stokes' Theorem, why can't you use a line integral? Stokes' Theorem just equates them. Using it either way is using the theorem. Yes?

Answer 47

http://www.math.ucsd.edu/programs/undergraduate/math_20E_exam/Practice_Test.pdf would be the link to the question also. It's the 3rd question.