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What is the best method of trying to find the surface integral of a scalar field over a certain surface $\Sigma$ that's only defined as being the part of one surface which is cut off by another surface?

E.g.

Find $\iint_\Sigma (x^4-y^4+y^2z^2-z^2x^2+1)d\sigma$ where $\Sigma$ is the part that the cylinder $x^2+y^2=2x$ cuts off of the top part of the cone $z^2=x^2+y^2$.

The way I tried solving it is by applying the following:

$$\iint_\Sigma f(x,y,z)d\sigma = \iint_K f(x(u,v),y(u,v),z(u,v))||\frac{\partial \phi}{\partial u}\times \frac{\partial \phi}{\partial v}(u,v)|| dudv$$

Where $\phi(K)=\Sigma$ is the parametrization of the surface:
$x=\phi_1(u,v)=u$
$y=\phi_2(u,v)=v$
$z=\phi_3(u,v)=u^2+v^2$
I then integrate from $0$ to $+\sqrt{2u-u^2}$ for $v$ and then from $0$ to $2$ for $u$. (This gives us half of what we need to find, the other half has an integration to the negative square root for $v$ but is left out because this is just an example.)

Now what I get is a very big integral:

$$\int_0^2du\int_0^\sqrt{2u-u^2} (u^4-v^4+v^2(u^2+v^2)^2-u^2(u^2+v^2)^2+1)\sqrt{4(u^2+v^2)+1}\ dv$$

This seems very complex to me which made me think that, although my method may be correct (which I'm not sure it is), it might be circuitous.

Is there a better way to go about solving these types of integrals?

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I would advise parameterizing $\Sigma$ with respect to the cylindrical coordinates $r$ and $\theta$. So

  • $x = r\cos\theta$
  • $y = r\sin\theta$
  • $z = r$

and the domain $K$ of the parameterization is defined by the inequalities $-\pi/2\leq \theta\leq \pi/2$ and $0\leq r\leq 2\cos\theta$. The surface area element $d\sigma$ in this case should be $$d\sigma = \left\|\frac{\partial\phi}{\partial r}\times \frac{\partial \phi}{\partial \theta}\right\|drd\theta = r\sqrt{2}drd\theta,$$ so that the integral you are trying to compute is far more reasonable: $$ \iint_\Sigma (x^4-y^4 + y^2z^2-z^2x^2 + 1)d\sigma$$$$ = \int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}(r^4\cos^4\theta-r^4\sin^4\theta+r^4\sin^2\theta-r^4\cos^2\theta + 1)r\sqrt{2}drd\theta.$$ While this may look ugly, after applying the trig identity $\sin^2\theta + \cos^2\theta = 1$ a couple of times, the first four terms in the integrand actually cancel out, and you're left with $$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}r\sqrt{2}drd\theta$$

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  • $\begingroup$ Right, it seems so obvious now that I should've used cylindrical coordinates for something like this considering the radial symmetry... Thanks! $\endgroup$ – Joshua May 23 '15 at 13:43

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