Surface Integral

**angel.white** · July 31st 2008, 12:48 AM

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

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

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

**mr fantastic** · July 31st 2008, 01:45 AM

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.

**angel.white** · July 31st 2008, 02:07 AM

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?

**mr fantastic** · July 31st 2008, 02:13 AM

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) .......

**angel.white** · July 31st 2008, 02:16 AM

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.

**mr fantastic** · July 31st 2008, 02:26 AM

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

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