# Which is the correct way to compute this surface integral?

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

$$x=f(y,z)$$

$$||n||=\sqrt{(f_y)^2+(f_z)^2+1}$$

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

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

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)