I am preparing for the GRE math subject test. And I am not sure whether my solution to this surface integral problem is correct.
Problem
- Surface $S$ is part of $x^2 +y^2 = 1$ between planes $z = 0$ and $x+y+z = 2$, a vector field $\vec F = x\vec i +y\vec j+z\vec k$, what is the value of integration $$\iint\limits_S\vec F\cdot \vec n\,\textrm dS,$$ where $\vec n$ is the unit normal vector of $S$?
My answer
Note that $\vec{n} = (x, y)$, \begin{equation} \iint_S \vec{F} \vec{n} dS = \iint_S (x^2+y^2) dS = \iint_S dS \end{equation} To evaluate the area of $S$, let $x= \cos\theta$ and $y= \sin\theta$, then \begin{equation} \vec{r} = (\cos\theta, \sin\theta, z) \end{equation} hence \begin{equation} \iint_S dS = \int_{0}^{2\pi} d\theta \int_0^{2-\cos\theta-\sin\theta} dz = 4\pi \end{equation}
Thanks in advance for pointing my mistakes if any.