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I need to calculate the surface integral of $xy$ over the surface $y^2+z^2=36$ in the first quadrant contained within $x^2+y^2=25$.

I started off trying to take a shortcut by solving $y^2+z^2=36$ for $z$: $z=\sqrt{36-y^2}$. While I think this is not incorrect, it gave an integral that was too difficult to solve. If it was incorrect to do this in the first place, let me know why.

So I took the standard approach and parameterized the equation using cylindrical coordinates, as the equation is that of a cylinder. The parameterization I chose was $$r=<x, 6\cos(\theta), 6\sin(\theta)>$$ so, $$r_x=<1, 0, 0>, r_\theta=<0, -r\sin(\theta), r\cos(\theta)>$$ and $$r_\theta \times r_x=<0, -r\cos(\theta), -r\sin(\theta)>$$ $$|r_\theta \times r_x |=6$$

Therefore, I need to integrate $6\int\int_D{xy}\ dA$. Now to integrate this, do I parameterize x and y in the same way as before? So $x=x$ and $y=6\cos(\theta)$? Or can I choose a new parameterization like $x=5\cos(\theta)$ or $y=5\sin(\theta)$? I'm assuming I can't use a different parameterization, and if not, then what would the domain of integration be? $\theta$ would range from $0$ to $\pi/2$ since it's in the first quadrant, but what would I choose for $x$? If I can choose a different parameterization, why can I?

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up vote 2 down vote accepted

Call the surface element ${\rm d}\omega$. You already have computed ${\rm d}\omega=6{\rm d}(x,\theta)$, and your aim is the integral $$I:=\int_S x\, y\ {\rm d}\omega=36 \int_D x\,\cos\theta\ {\rm d}(x,\theta)\ ,$$ where $S$ denotes the intended piece of surface and $D$ is the corresponding domain in the $(x,\theta)$ parameter plane. To determine $D$ we have to assure $x\geq0$, $y\geq0$, $z\geq0$, and $x^2+y^2\leq 25$ for all points $(x,y,z)\in S$. So a priori $x\geq0$ and $0\leq\theta\leq{\pi\over2}$. Furthermore there is the restriction $x^2+36\cos^2\theta\leq25$ which leads to $0\leq\cos\theta\leq{\sqrt{25-x^2}\over6}$. Therefore we definitively have $$D=\Bigl\{(x,\theta)\ \bigm|\ 0\leq x\leq 5,\ \arccos{\sqrt{25-x^2}\over6}\leq\theta\leq{\pi\over2}\bigr\}\ .$$ (It would help to draw a three-dimensional figure showing $S$ and $D$!).

Our integral $I$ then becomes $$I=36\int_0^5\int_{\arccos{\sqrt{25-x^2}\over6}}^{\pi/2} x \cos\theta\ d\theta \ dx\ .$$ The inner integral is $$\int_{\arccos{\sqrt{25-x^2}\over6}}^{\pi/2} \cos\theta\ d\theta =\sin\theta\Bigr|_{\arccos{\sqrt{25-x^2}\over6}}^{\pi/2} =1-{1\over6}\sqrt{11+x^2}\ .$$ So it remains to compute $$I=36\int_0^5 x\Bigl(1-{1\over6}\sqrt{11+x^2}\Bigr)\ dx\ ,$$ which I may leave to you.

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I might be missing something obvious here, but how did you figure out that $\cos(\theta)\le \frac{5}{6}$? – gsingh2011 Apr 10 '12 at 15:07
Oh, I think I see. The maximum value of y is when x=0. Then $0+y^2=25$, so $6\cos(\theta)=5$. Thanks for the answer, I'll get back to you after I complete the problem. – gsingh2011 Apr 10 '12 at 15:11
So first of all, your upper bound for x is always negative, so it should really be a lower bound. But x should never go below zero because it's in the first quadrant... – gsingh2011 Apr 10 '12 at 23:16

The area density on the surface at the point above $(x,y,0)$ will be $f(y) dA$ for some function of $y$ and you might be able to integrate $xyf(y) dA$ over the interior of the circle of radius 5. Here $dA$ is area measure in the $xy$ plane and $df$ is arclength on the circle of radius 6.

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