Question

determine the surface area of the surface given parametrically by r(u,v) = , 0 is less than or equal to u which is less than or equal to 1, and 0 is less than or equal v which is less than or equal to 1.

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

i surface integral is similar to a line integral ....murmur

Answer 3

i spose it begins as the sqrt([dx]^2 + [dy]^2 + [dz]^2) right?

Answer 4

it's a step somewhere.

Answer 5

wish i could do more; but thats as far as I can get in this one :)

Answer 6

at least the bounds are constants :)

Answer 7

x= u+v
y = v
z = u

x = z+y maybe?

x' = z' + y'

Answer 8

your gonna have to hope a smarter then me comes along ;)

Answer 9

or at least chimes in ;)

Answer 10

it's no problem. thanks for trying tho. there is a question you might can answer for me. when using a determinant to find the cross product of 2 vectors, which way do you cross multiply first? bottom left to top right or bottom right to top left?

Answer 11

its just the nature of the best; the matrix itself is just a way to organize the information; but the results of acutally going thru the work results in the technique

Answer 12

the crossing of vectors mimics the determinate of a matrix; so its a good visual to follow

Answer 13

but spose we have 2 vectors on a plane froma single point and we want to determine the normal to the plane... we would set up the proper dots products to get a zero from each vector and the resulting summations result in a form tha tmatches the determinate of a matrix...

Answer 14

its alot to try to type in here :) but if you need more clarification, just let me know

Answer 15

I think it's
\[\int\int | {\delta r \over \delta x} \times {\delta r \over \delta y}|dxdy\]
Where r is what you have defined there parametrically (so you'll probably need a Jacobian as well).

Answer 16

isn't the cross product the jacobian?

Answer 17

your equation is correct according to what i just found.

Answer 18

Oh.. I'm sorta right.. This should help some:
http://www.math24.net/surface-integrals-of-first-kind.html

Answer 19

thanks for the link. i found the answer i needed. I found the definition of the surface area.

Answer 20

In particular this seems promising:

\[\int\int_SdS = \int\int_{D(u,v)}r(u,v)| {\delta r \over \delta u} \times {\delta r \over \delta v}|dudv\]

Answer 21

that's what i found. thanks for your help.