Question

Evaluate the surface integral

z dS where S is the part of the plane 4x+3y+z=12 that lies in the first octant.

Answer 1

\[\int\limits_{?}^{?} z DS\]

Answer 2

@Phi did one similar a while back, here's the link. http://assets.openstudy.com/updates/attachments/514bb348e4b05e69bface25f-phi-1364172797512-evaluatethesurfaceintegral.pdf

Answer 3

@mathmale

Answer 4

For a scalar function:
\[\int\limits_{S}^{} f(x,y,z) dS = \int\limits_{R}^{} \int\limits_{}^{} f(x(u,v),y(u,v),z(u,v))*|r_{u} \times r_{v}| dA\]

Answer 5

You have to rewrite the surface in parametric form.
Sorry, if I'm no help. ;-;

Answer 6

rewrite the equation of the surface*

Answer 7

so z=12-4x-3y parametizes the surface...

Answer 8

r(u,v) = <u, v, 12 - 4u - 3v>

Answer 9

You can rewrite the function in terms of u and v now, or do that after the cross product.

Answer 10

right or (x,y, 12-4x-3y)

Answer 11

Uh, sure. Find the partial derivative of the function wrt u and v.

Answer 12

Um, Idk what you did there.

Answer 13

F wrt u=-4, wrt v=-3

Answer 14

those are the partials^

Answer 15

No the partial derivative of the vector function:
r(x,y) = <x, y, 12 - 3x- 4y>
With respect to x and y.

Answer 16

We need to cross product rx and ry. Then find the magnitude.

Answer 17

idk what you are saying?

Answer 18

I know. Calculus sucks in typing.

Answer 19

OH nvm. ok so <1,1, 12-3x-4y> right?

Answer 20

No, we need partial derivative of x and y separately.
rx = <1, 0, -3>

Answer 21

Your (x,y, 12-4x-3y) appeals to me because it expresses three variables in terms of just two: x and y.

Unfortunately, it's been some months since I last did problems of this nature, and I'd have to review that material to be of significant help. iPwnBunnies seems to be current and well informed, so I'd encourage you two to continue working together on this problem.

Answer 22

Yes, it's a parametric form of the surface lol. Since for the surface, z is a function of x and y.

Answer 23

Need it for the working definition of a surface integral.

Answer 24

oh sorry I didn't understand what you were asking for

Answer 25

ry= <0,1,-4>

Answer 26

Oh crap, I think I mixed up the coefficients for the equation of the surface.

Answer 27

yeah you did. but fixing it would be <1,0,-4> and <0, 1, -3>

Answer 28

for rx and ry respectively

Answer 29

z = 12 - 4x - 3y
r(x, y) = <x, y, 12 - 4x - 3y>
rx = <1,0,-4>
ry = <0, 1, -3>

Good. Can you cross product rx and ry, and find the magnitude?

Answer 30

<4, -3, 1> magnitude of sqrt(16+9+1) or sqrt(26)

Answer 31

Good, let's go back to the working definition. I'll just draw it.

Answer 32

We have the cross product done. Our function is not in terms of x and y yet doh. We can get it from the parametric form of the surface though.