1. ## Surface Integral

Evaluate the surface integral:
$\int \int_S f(x,y,z) ~dS$

S is the surface with parametric equations $x=u^2$, $y=u~sin~v$, $z = u~cos~v$

$0\leq u\leq1$
$0\leq v\leq \frac{\pi}2$

2. Originally Posted by angel.white
Evaluate the surface integral:
$\int \int_S f(x,y,z) ~dS$

S is the surface with parametric equations $x=u^2$, $y=u~sin~v$, $z = u~cos~v$

$0\leq u\leq1$
$0\leq v\leq \frac{\pi}2$
$dS = \left|\vec{dS} \right| = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right| du \, dv$ where $\vec{r} = x(u, v) \, i + y(u, v) \, j + z(u, v) \, k$.

For your question, $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = -u \, i + 2u^2 \sin v \, j + 2u^2 \cos v \, k$.

Therefore $dS = u \sqrt{1 + 4u^2} \, du \, dv$.

So your surface integral becomes $\int_{v=0}^{\pi/2} \int_{u=0}^{1} g(u, v) \, u \sqrt{1 + 4u^2} \, du \, dv$ where g(u, v) is what you get from f(x, y, z) after making the parametric substitution.

3. Originally Posted by mr fantastic
$dS = \left|\vec{dS} \right| = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right| du \, dv$ where $\vec{r} = x(u, v) \, i + y(u, v) \, j + z(u, v) \, k$.

For your question, $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = -u \, i + 2u^2 \sin v \, j + 2u^2 \cos v \, k$.

Therefore $dS = u \sqrt{1 + 4u^2} \, du \, dv$.

So your surface integral becomes $\int_{v=0}^{\pi/2} \int_{u=0}^{1} u \sqrt{1 + 4u^2} \, du \, dv$.
Okay, I follow you this far, if f(x,y,z) = yz then I should be finding:

$\int_{v=0}^{\pi/2} \int_{u=0}^{1} u^3\sin v \cos v \sqrt{1 + 4u^2} \, du \, dv$

= $\int_{0}^{\pi/2} \sin v \cos v \, dv \int_{0}^{1} u^3 \sqrt{1 + 4u^2} \, du$

Is this correct?

4. Originally Posted by angel.white
Okay, I follow you this far, if f(x,y,z) = yz then I should be finding:

$\int_{v=0}^{\pi/2} \int_{u=0}^{1} u^3\sin v \cos v \sqrt{1 + 4u^2} \, du \, dv$

= $\int_{0}^{\pi/2} \sin v \cos v \, dv \int_{0}^{1} u^3 \sqrt{1 + 4u^2} \, du$

Is this correct?
Yes. And you know the substitution to make with the u-integral ....?

By the way, I edited my earlier reply. I'm glad to see you realised I forgot the f(x, y, z) .......

5. Originally Posted by mr fantastic
Yes. And you know the substitution to make with the u-integral ....?

By the way, I edited my earlier reply. I'm glad to see you realised I forgot the f(x, y, z) .......
Right now I'm trying $a^2 = 1+4u^2$

I have gotten this far several times, but seem to keep integrating wrong. I needed to verify that I was correct up to this point so I could zero in on my error.

6. Originally Posted by angel.white
Right now I'm trying $a^2 = 1+4u^2$

I have gotten this far several times, but seem to keep integrating wrong. I needed to verify that I was correct up to this point so I could zero in on my error.
Try $w = 1 + 4u^2$.

You should get $\frac{5\sqrt{5}}{24} + \frac{1}{120} = \frac{25\sqrt{5} + 1}{120}$.