I'm not seeing what i did wrong here for this vector calculus problem. If anyone could point me in the right direction i would be most appreciative. The problem reads:
Let $$ F= (2yz, -x+3y+2,x^2 +z) $$ Evaluate $$\iint_{S}^{} \operatorname{curl}(F)\cdot dS$$ where $S$ is the cylinder $$ x^2 + y^2 = a^2 $$ for $$0\leq z\leq 1$$ without the top and bottom. What if the top and bottom are included?
My attempt:
Let $$ c(t) = (a \cos(t) , a \sin(t), 0) $$ $$ c'(t) = (-a \sin(t) , a \cos(t), 0) $$ $$ F(c(t)) = (0, -a \cos(t) + 3a \sin(t) + 2, a^2 (\cos(t))^2) $$ So, $$ \iint_{s} \operatorname{curl}F \cdot dS = \int_{0}^{2 \pi} (0, -a\cos(t)+3a\sin(t)+2, (acos(t))^2)\cdot(-a\sin(t), a\cos(t),0)dt $$ $$=\int_{0}^{2\pi}-(a\cos(t))^2 + 3a^2\sin(t)\cos(t)+2a\cos(t))dt$$ $$=-a^2\pi $$
The answer should be $$ I = 2a^2\pi $$
For the 2nd part I see that there is no boundary being that it is now a closed surface and the integral is zero.
