1. ## Vector Calculus

Hi.

Calculate the integral:

$\iint\limits_S\ f(x,y,z)dS$

f(x, y, z) = xy

r(u, v) = (u - v, u + v, 2u + v + 1) u[0, 1] ; v[0, u]

What i didn't get is how to calculate the limits of the double integral, how to transform v[0, u] in v[0, number].

A.: $\sqrt {14} / 6$

______________________________

I've tried to calculate in this way (but it didn't worked)

A = $\int_0^u \int_0^1 (u^2 - v^2) \left ||{{\vec{r_u}.\vec{r_v}}} \right || dudv$

2. ## Re: Vector Calculus

You need to know the region in the $(u, v)$-plane that you're integrating over. This should be specified in the problem. What you've written, $u[0,1];v[0,u]$, doesn't make any sense.

3. ## Re: Vector Calculus

Originally Posted by PedroMinsk
Hi.

Calculate the integral:

$\iint\limits_S\ f(x,y,z)dS$

f(x, y, z) = xy

r(u, v) = (u - v, u + v, 2u + v + 1) u[0, 1] ; v[0, u]

What i didn't get is how to calculate the limits of the double integral, how to transform v[0, u] in v[0, number].

A.: $\sqrt {14} / 6$

______________________________

I've tried to calculate in this way (but it didn't worked)

A = $\int_0^u \int_0^1 (u^2 - v^2) \left ||{{\vec{r_u}.\vec{r_v}}} \right || dudv$
As ojones said, this makes no sense. The "outside" integral has an upper limit of "u" and so the integral will be a function of u, not a constant. Did you mean
$\int_{u= 0}^1\int_{v= 0}^u (u^2- v^2) ||\vec{r_u}\times\vec{r_v}||dudv$?
(not dot product!)

That should work.

4. ## Re: Vector Calculus

I'm sorry. I solved it now. It was so simple...

Thanks.