# Math Help - Surface Integrals

1. ## Surface Integrals

Hi,

is the formula for ds equal to

$
ds = sqrt(( (pz/px)^2 + (pz/py)^2 + 1))$

where p = partial derivative

many thanks

Problem:
For the parametrically defined surface S given by , find each of the following differentials.

In , dS = ________ du dv In , dS = ___________ du dv

2. Originally Posted by althaemenes
Hi,

is the formula for ds equal to

$
ds = sqrt(( (pz/px)^2 + (pz/py)^2 + 1))$

where p = partial derivative

many thanks

Problem:
For the parametrically defined surface S given by , find each of the following differentials.

In , dS = ________ du dv In , dS = ___________ du dv

$d\vec S =\vec n \cdot dS=\frac{(\vec r_u \times\vec r_v)}{|\vec r_u \times \vec r_v|} |\vec r_u \times \vec r_v|dA=(\vec r_u \times \vec r_v)dA$

3. Originally Posted by TheEmptySet
$d\vec S =\vec n \cdot dS=\frac{(\vec r_u \times\vec r_v)}{|\vec r_u \times \vec r_v|} |\vec r_u \times \vec r_v|dA=(\vec r_u \times \vec r_v)dA$
okay, So I got <0,0,0> for the first ds and 0 for the second ds

Given,

$
r_u=-vsin(uv)i + vcos(uv)j + 0k

$

$
r_v=-usin(uv)i + ucos(uv)j + 0k

$

$(\vec r_u \times \vec r_v)dA = 0dA$

but its wrong!!!

4. ## sorry!!!

I did it wrong the first time....

Now I got it... Thanks a lot ..... YOU ROCK

5. I am working on the same question. I got the answer to the first part:

<vcos(uv),vsin(uv),0>

but I cant figure out the answer to the second part. Any help would be greatly appreciated.

6. ## reply

for the second part the answer is: v