I am trying to find the Surface integral of the surface defined as the part of the cylinder $x \le 0$, $x^2+y^2=1$, $x \le 0$ and $0 \le z\le -\sqrt{3}x$ and vector field $\vec{F}=x\hat{i}+y\hat{j}$
I first found out the normal vector of cylindrical surface given by:
$$\vec{n}=2x\hat{i}+2y\hat{j}$$
So $$\hat{n}=x\hat{i}+y\hat{j}$$
So $\vec{F}.\hat{n}=x^2+y^2=1$
So the surface integral is:
$$I=\int\int_{S}1.dS$$
Now how to parameterize this surface?
It is convenient to change to cyllindrical coordinates (with $r=1$): $$x = \cos\theta \quad y = \sin\theta \quad z=z$$ Now, you gonna have to integrate over $\theta$ and $z$. You have: $$\sqrt{3}\cos\theta \le z \le 0 \quad \mbox{and} \quad \frac{\pi}{2} \le x \le \frac{3\pi}{2}$$ Thus: $$\int\int_{S}1dS = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\int_{-\sqrt{3}\cos\theta}^{0}1dzd\theta$$