Calculate the flow of $\vec{F}(x,y,z) = 3x\vec{i} + 3y\vec{j} + z^5\vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 \leq z \leq 1$ Calculate the flow of $\vec{F}(x,y,z) = 3x\vec{i} + 3y\vec{j} + z^5\vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 \leq z \leq 1$. The normal considered points inside.
The book uses cylindrical coordinates and the final answer is $-150 \pi$. I want to see other ways to approach the problem so I could understand better the concept. Must we not use the jacobian in this case? How would I solve the problem with cartesian coordinates?
 A: First parametrize the surface  $S$, $\vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=\pm\sqrt{25-t^2}$ and $z=s$ free:
$$\vec r=(\pm\sqrt{25-t^2}, t,s)\;;-5\leq t<5\;;0\leq s\leq 1$$
$\vec  F$ on the surface is $\vec F=(\pm3\sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$\dfrac{\partial \vec r}{\partial t}=\left(\dfrac{\mp t}{\sqrt{25-t^2}},1,0\right)$
$\dfrac{\partial \vec r}{\partial s}=(0,0,1)$
$\dfrac{\partial \vec r}{\partial t}\times\dfrac{\partial \vec r}{\partial s}=\left(1,\dfrac{\pm t}{\sqrt{25-t^2}},0\right)$
Then, the flux,
$$\Phi=\int_S\vec F·\left(\dfrac{\partial \vec r}{\partial t}\times\dfrac{\partial \vec r}{\partial s}\right)\;\mathbb d t\,\mathbb ds$$
$$\Phi=\int_{0}^1\int_{-5}^5\left(3\sqrt{25-t^2}+\dfrac{3t^2}{\sqrt{25-t^2}}\right)\,\mathbb d t\,\mathbb ds+\int_{0}^1\int_{5}^{-5}\left(-3\sqrt{25-t^2}-\dfrac{3t^2}{\sqrt{25-t^2}}\right)\,\mathbb d t\,\mathbb ds$$
$$\Phi=150\pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $\vec F$ points outwards too on the surface)
