Let $\overrightarrow {F} : \Bbb R^3 \longrightarrow \Bbb R^3$ be a vector field defined by $F(x,y,z)=(e^y,0,e^x),\ (x,y,z) \in \Bbb R^3$. Let $S$ be the surface given by $S = \left \{(x,y,z) \in \Bbb R^3 : x^2+y^2 = 25, x \geq 0 , 0 \leq z \leq 2 \right \}.$ Then find the following surface integral $$\iint_S \overrightarrow {F} \cdot d \overrightarrow {A}.$$
My attempt $:$ If I use Gauss' divergence theorem then the integral clearly evaluates to $0$ as because $\operatorname {div} \left (\overrightarrow F \right ) = 0.$
Now let us compute the same integral from the definition of surface integral. For this we first observe that the given surface is cylindrical in shape. So it is better to change the cartesian coordinates into cylindrical coordinates. Then the given surface can be written in terms of the graph of a function known as the parameterization of the surface. Consider the function $\varphi : \left [-\frac {\pi} {2} , \frac {\pi} {2} \right ] \times [0,2] \longrightarrow \Bbb R^3$ defined by $$\varphi (u,v) = (5 \cos u, 5 \sin u, v),\ u \in \left [-\frac {\pi} {2}, \frac {\pi} {2} \right ], v \in [0,2].$$ Then it is easy to see that the graph of $\varphi$ is the given surface $S$ i.e. $\varphi$ is a parameterization of $S.$ Now $\varphi_u = (-5 \sin u, 5 \cos u, 0)$ and $\varphi_v = (0,0,1).$ So $\varphi_u \times \varphi_v = (5 \cos u, 5 \sin u, 0).$ So the unit normal vector $\hat {n}$ at any point $(u,v)$ is given by $\frac {\varphi_u \times \varphi_v} {\left \| \varphi_u \times \varphi_v \right \|}.$ Now with these things in mind let us compute the given surface integral. $$\begin{align*} \iint_S \overrightarrow {F} \cdot d \overrightarrow {A} & = \iint_S \overrightarrow {F} \cdot \hat {n}\ dA \\ & = \int_{0}^{2} \int_{-\frac {\pi} {2}}^{\frac {\pi} {2}} \overrightarrow {F} \cdot \frac {\varphi_u \times \varphi_v} {{\left \| \varphi_u \times \varphi_v \right \|}} {{\left \| \varphi_u \times \varphi_v \right \|}}\ du\ dv \\ & = 5 \int_{0}^{2} \int_{-\frac {\pi} {2}}^{\frac {\pi} {2}} e^{5 \sin u} \cos u\ du\ dv \\ & = 10 \int_{-\frac {\pi} {2}}^{\frac {\pi} {2}} e^{5 \sin u} \cos u\ du \\ & = 10 \int_{-1}^{1} e^{5t}\ dt \\ & = 2(e^5 – e^{-5}) \neq 0. \end{align*}$$
What's going wrong in my calculations? Any suggestion regarding this will be highly appreciated.
Thank you very much for your valuable time.
Best Answer
Your computation looks fine. What's definitely wrong is trying to apply Gauss' Theorem to a surface that is not closed.