# Homework Help: A surface integral

1. Dec 15, 2012

### BurningHot

1. The problem statement, all variables and given/known data
Compute the integral
$\iint_S \sin y dS$
where S is part of the surface $x^2 +z^2 = \cos^2(y)$ lying between the planes $y=0$ and $y=\pi/2$.

2. Relevant equations
$\iint_S f(x,y,z) dS = \iint_D f(x,y, g(x,y)) \sqrt{g_x^2 +g_y^2 +1}dA$
$\iint_S f(x,y,z) dS = \iint_D f(\vec{r}(u,v)) \|\vec{r}_v\times\vec{r}_v\| dA$

3. The attempt at a solution
So my attempt was to parametrize to surface as $x=x , y=y, z=\sqrt{\cos^2y - x^2}$, then put everything in the equation, and ended up with this integral $\int_0^{\pi/2}\int_0^{cosy} siny \sqrt{\frac{x^2+sin^2ycos^2y}{cos^2y-x^2} +1} dxdy$

This is exactly what I would get if I took the partial derivatives and used the other equation. I don't think I've made a mistake in my calculations, but that's more than likely to happen.
I'm not sure how to proceed from here. Is there some kind of factorization I can do? Some identity to apply? Some substitution to use?

2. Dec 15, 2012

### haruspex

I've never worked in this area, so I may have this completely wrong, but I tried r = (x, sin(y), z), taking x and z as the parameters. This simplified nicely and I ended up with pi/4. Is that valid, or must I use r = (x, y, z)?

3. Dec 15, 2012

### BurningHot

How did you end up with r = (x, sin(y), z)? If you take x and z to be the parameters, wouldn't y = cos-1√(x^2+z^2) ?

When parameterizing a surface, you use two variables, right?
With what I did, my surface ended up being parametrized as $\vec{r}(x,y) = <x, y, \sqrt{cos^2y-x^2}>$

4. Dec 15, 2012

### haruspex

I was not happy with my r = (x, sin(y), z), so I tried again with the more convincing r = (x, y, z). Taking x and z as parameters, r = (x, √(1-x2-z2), z). ||rx x rz|| = 1/sin(y), so the integral reduces to ∫∫dxdz. All that's left is to be careful with the ranges in the double integral. (This is the same result as I got with r = (x, sin(y), z), so maybe it was valid.)

5. Dec 15, 2012

### BurningHot

Hm... ok. I think I see what's happening now.
Thank a lot for the help.