I am trying to find a surface integral $$\iint_Syz\ dS$$ of a cylinder segment where $S$ is the portion of $x^2 + y^2 = 1$ with $x ≥ 0$ and $z$ between $z = 2$ and $z = 5 − y$.

I thought that there should be two ways to do this. The first is to use rectangular coordinates, and do this



$$={1\over \sqrt{1-y^2}}$$

$$\int_{-1}^1\int_2^{5-y}\ {yz\over \sqrt{1-y^2}} \, dz \, dy =-\frac{5\pi}{2} \approx -7.854$$

I also tried parameterizing the cylinder with

$$r_z(z,\theta)=\left<0,0,1\right>$$ $$r_{\theta}(z,\theta)=\left<-\sin\theta,\cos\theta,0\right>$$ $$||n||=\sqrt{(-\sin\theta)^2+(\cos\theta)^2+0^2}=1$$ $$\int_{0}^\pi\int_2^{5-\sin\theta}\ z\sin\theta \, dz \, d\theta \approx 13.813$$

Why am I getting different solutions? I though it might have to do with the fact that the first approach is taking into account the negative parts of $y$, and indeed when I circumvent this I get the same solution as the parametrization:

$$2\int_{0}^1\int_2^{5-y}\ {yz\over \sqrt{1-y^2}} \, dz \, dy \approx 13.813$$

But it is still unclear to me which is the correct approach, and why.


Your limits of $\theta$ are wrong in the second integral. Because you want $x \geq 0$, you need to integrate over $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ instead of $0 \leq \theta \leq \pi$.

Indeed, $$ \int_{-\pi/2}^{\pi/2} \int_0^{5-\sin\theta} z \sin\theta\,dz\,d\theta = -\frac{5\pi}{2} \approx -7.854 $$ (according to Wolfram Alpha)


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