Question

evaluate the surface integral

Answer 1

\[\int\limits_{}^{}\int\limits_{s}^{}(xy) dx \lambda dz-(yz) dz \lambda dx + (xz) dx \lambda dy\]
where S is the part of the plane x + y + z = 1 lying in the rst octant. Use x and y
as parameters.

Answer 2

note* I couldn't find a wedge so I used lamba instead for the jacobian

Answer 3

you for the exterior product? \( dx \wedge dz \) etc?

Answer 4

I'm thinking that for the first part, ∂(y,z)/∂(x,y) = \[\left[\begin{matrix}∂y/∂x & ∂z/∂x \\ ∂y/∂y & ∂z/∂y\end{matrix}\right]\]

Answer 5

= (∂y/∂x) (∂z/∂y) - (∂y/∂y) (∂z/∂x)

Answer 6

yes james that's what i meant

Answer 7

hmm. As you've articulated, the problem doesn't quite make sense. What's the original problem or motivation for this question?

Answer 8

wait. Is it the Jacobian or is it the exterior product?

Answer 9

it's just a homework assignment on it's own

Answer 10

it's the jacobian...don't think we learned exterior product

Answer 11

I don't under why you want the Jacobian at all. Are x,y,z just the regular coordinate variables here, or are they truly variables of the other function?

And even if it is the Jacobian, it doesn't make sense that it is placed between the dx, dy, dz terms.

Answer 12

i just thought that dy∧dz meant the jacobian...?

Answer 13

No. It means something else. So you do mean dy∧dz etc.? These are called exterior products.

Answer 14

yeah that's what i mean...but I don't have anything about exterior product in my notes...I don't think that's what the prof had in mind

Answer 15

my notes say: dy∧dz = ∂(Y,Z) / ∂(u,v)

Answer 16

It cannot be the case that dy∧dz = ∂(Y,Z) / ∂(u,v) because if it did, then were the d(somethings) for integration?

What is true is that

dy∧dz = ∂(Y,Z) / ∂(u,v) du∧dv

But we don't need to change variables right now, so the Jacobian part of this is irrelevant.

So we are integrating (the 2-form)
\[ F = (xy)\ dx \wedge dz−(yz\ )dz \wedge dx+(xz)\ dx \wedge dy \]
over the surface S.

Now are you sure about this F. Should the second term by \( -(yz) dz \wedge dy \)?

Answer 17

Correction: It cannot be the case that dy∧dz = ∂(Y,Z) / ∂(u,v) because if it did, then where are the d(somethings) for integration?

Answer 18

1) yes I'm sure that's the question
2) dy∧dz = ∂(Y,Z) / ∂(u,v) dudv

Answer 19

and since we are using x and y as parameters I thought that it would mean dy∧dz = ∂(Y,Z) / ∂(x,y) dxdy

Answer 20

With x and y as parameters,
z = 1 - x - y
... you can use your Jacobian here when you have a dz term.

and the limits of integration are
y: 0 to 1 - x
x: 0 to 1

Answer 21

okay so I have this determinant...how do I find ∂y/∂x?
[∂y/∂x ∂y/∂y]
[∂z/∂x ∂z/∂y]

Answer 22

y = 1 - x - z

∂y/∂x = 0 - 1 - 0
∂y/∂x = -1 ?

Answer 23

You're asked to use x and y as parameters; that's why I wrote
z = 1 - x - y.

The only variable you're changing here is the z. So you don't need to calculate terms such as ∂y/∂x

Answer 24

so then
[∂y/∂x ∂y/∂y]
[∂z/∂x ∂z/∂y]
=
[ 0 0 ]
[∂z/∂x ∂z/∂y]
= 0?

Answer 25

I've never seen lambda at all used in this way. What is lambda in the first post, in this context?