# Problem calculating an integral over a surface

I've been trying to solve this for awhile and can't find a way. Given $S={(x,y,z) \in R^3 : z = x^2 - y^2 , x^2 + y^2 \leq 1 }$ and $\phi :R^3 \to R$ defined as $\phi (x,y,z)= (4z +8y^2 + 1)^{3/2}$, find $\iint_{S} \phi dS$.

I think I have to use $\int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta)) \left \| T_{\theta} \times T_{r} \right \| dr d\theta$ So I first parametrized S with $x= r \cos \theta$, $y= r \sin \theta$ and $z = r [2\cos^2 \theta - 1]$. So I had $T(r,\theta)= (r\cos\theta , r\sin\theta, r[\cos^2\theta - 1])$.

If I'm not mistaken I have to use $\int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta) \left \| T_{\theta} \times T_{r} \right \| dr d\theta$.

Now I need to find $|| T_{r} \times T_{\theta} || = || (r\cos\theta[2\cos^2 \theta-1+2\sin\theta], r\sin\theta, -r) ||$

Now, as you can imagine that ends up being one messed up thing once you square the first term, and I haven't found anyway to simplify anything, which leaves me with a pretty tough integral to solve.

This is the integral I got, but Wolfram Alpha couldn't solve it and neither could I.

$\int_{0}^{2\pi} \int_{0}^{1} (8 r^2 \sin^2(\theta)+4 r(2 \cos^2(\theta)-1)+1)^{3/2} \sqrt((r \sin(\theta))^2+(2 r \cos^3(\theta)+2 r \sin^2(\theta) \cos(\theta)-r \cos(\theta))^2+(-r \cos^2(\theta)-r \sin^2(\theta)|^2) dr d\theta$

I'm guessing I either used a wrong parametrization, or there's another way to solve this not using $\iint_{S} \phi(T(r,\theta)) \left || T_{r} \times T_{\theta} \right || dS$.

Any help would be greatly appreciated.

EDIT: As Didier pointed out, $z = r^2 [2 \cos^2 \theta +1]$. But that doesn't make matters a whole lot simpler since now $|| T_{r} \times T_{\theta} || = || (-4r^2 \cos \theta \sin^2 \theta - 2 r^2 \cos\theta [2\cos^2 \theta +1], -4 r^{2} \cos^{2} \theta \sin \theta + 2r^2 \sin \theta [2\cos^2\theta +1], r) ||$.

EDIT2: Another retarded mistake, fixed by Didier. $|| T_{r} \times T_{\theta} || = (-4r^2 \cos\theta\sin^2\theta-2r^2 \cos\theta[2\cos^2 \theta -1], 4r^2 \cos^2 \theta \sin\theta-2r^2\sin\theta[2\cos^2\theta-1], r)$, still not totally right though.

I must have another error somewhere but just can't seem to find it.

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Note that $z=r^2(2\cos^2\theta-1)$ and not $z=r(2\cos^2\theta-1)$. The subsequent formulas should be much simpler. – Did May 13 '11 at 0:07
@Didier I checked it a couple of times, and still didn't catch that, I could really use some sleep. Thanks! – Bananas May 13 '11 at 0:13
@Didier It still ends up in a pretty messed up integral. What makes things so complicated I think is that $\phi : \mathbb{R}^3 \to \mathbb{R}$, because if $\phi : \mathbb{R} \to \mathbb{R}^3$ I should be able to use $\iint_{S} (\phi . N) dS$, right? – Bananas May 13 '11 at 1:49
You also made a mistake writing $T(r,\theta)$. And your formula for $\|T_r\times T_\theta\|$ is wrong. My advice is to re-compute carefully $T$, $T_r$, $T_\theta$ and finally $\|T_r\times T_\theta\|$. – Did May 13 '11 at 1:49
@Didier Regarding T I made a mistake when copying, but its right (I think) on paper, $T(r,\theta)= (r\cos\theta, r\sin\theta, r^2 [2\cos^2\theta -1]$. I was doing $T_{r} \times T_{\theta}$ wrong though, I forgot that in the middle you multiply by -1, but it doesn't really affect the end product much, I still have $|| T_{r} \times T_{\theta} = (-4r^2 \cos\theta\sin^2\theta-2r^2 \cos\theta[2\cos^2 \theta -1], 4r^2 \cos^2 \theta \sin\theta-2r^2\sin\theta[2\cos^2\theta-1], r)$ – Bananas May 13 '11 at 2:18

I believe you're still looking for something to find your mistake with. Now your things aren't looking that bad from here so I guess just giving you the solution will let you see where you're wrong in your numbers crunching.

Let $S = \{ T(r,\theta) = (r \cos \theta, r \sin \theta, r^2 (\cos^2 \theta - \sin^2 \theta)) \, | \, \theta \in [0,2\pi], r \in [0,1] \}$ be the chosen parametrization of $S$. Therefore

$\phi(T(r,\theta)) = (4(r^2(\cos^2 \theta - \sin^2 \theta)) + 8 (r \sin \theta)^2 + 1)^{3/2}$

$= (4r^2 \cos^2 \theta - 4 r^2 \sin^2 \theta + 8 r^2 \sin^2 \theta + 1)^{3/2} = (4r^2 + 1)^{3/2}$

and

$T_r \times T_{\theta} = \begin{vmatrix} \vec i & \vec j & \vec k \\ \cos \theta & \sin \theta & 2r(\cos^2 \theta - \sin^2 \theta) \\ - r \sin \theta & r \cos \theta & r^2(-4 \sin \theta \cos \theta) \end{vmatrix}$

$= (-4r^2\sin^2 \theta \cos \theta - 2r^2 \cos^3 \theta + 2r^2 \sin^2 \theta \cos \theta, 4r^2 \sin \theta \cos^2 \theta - 2r^2 \sin \theta \cos^2 \theta + 2r^2 \sin^3 \theta, r)$

$=(-2r^2 \cos \theta, 2 r^2 \sin \theta,r) \quad$ so that $\quad ||T_r \times T_{\theta}|| = \sqrt{4r^4 + r^2} = r \sqrt{4r^2 + 1}$. Now it's easy because $$\int_0^{2\pi} \int_0^1 (4r^2+1)^{3/2} (r\sqrt{4r^2 +1}) \, dr d\theta = 2\pi \int_0^1 r(4r^2 + 1)^2 \, dr$$ which gives a number I am not in the mood for computing right now, but I think you'll be fine. =)

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Thanks. Apparently substituting $\cos^2 \theta - \sin^2 \theta = 2\cos^2\theta -1$ made everything more convoluted and I couldn't really catch on the simplifications. – Bananas May 13 '11 at 20:52
Using an identity you don't need to use at first is usually not a good idea. Keep things as they are as long as you don't see the need to change them. What I mean is that you can use identities to simplify things, but don't do it if it doesn't. Waste of time and effort. I'm glad it helped though. =) – Patrick Da Silva May 14 '11 at 4:22