Integrating $\nabla \times{F}$ - Finding the Error

[Telemachus]

Homework Statement

Hi there. I was trying to solve this problem, from the book. The problem statement says:
Integrate \nabla \times{F},F=(3y,-xz,-yz^2) over the portion of the surface 2z=x^2+y^2 under the plane z=2, directly and using Stokes theorem.

So I started solving the integral directly. For the rotational I got:

\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3)

Then I parametrized the surface.:

\begin{matrix}x=x=r\cos \theta\\ y=r \sin \theta \\z=\displaystyle\frac{r^2}{2}\end{matrix}, \theta[0,2\pi] ,r[0,2]

Then I did: T_r\times{T_{\theta}}

T_r=(\cos \theta,\sin \theta,r),T_{\theta}=(-r\sin \theta,r\cos \theta,0)

For the cross product I got:
T_r\times{T_{\theta}}=(r^2\cos \theta,-r^2\sin \theta,r)

And then I computed the integral

\displaystyle\int_{0}^{2}\int_{0}^{2\pi}(r\cos \theta-\displaystyle\frac{r^4}{4},0,\displaystyle\frac{-r^2}{2}-3)\cdot{(-r^2\cos \theta,-r^2\sin \theta,r)}d\theta dr=-12\pi

The result given by the book is: 20\pi.

I don't know what I did wrong, and is one of the first exercises that I've solved for the stokes theorem, so maybe I could get some advices and corrections from you :)

Thank you in advance. Bye.

 

[Telemachus]

I've corrected some typos. Is there anybody out there? :P

 

[HallsofIvy]

Homework Statement

Hi there. I was trying to solve this problem, from the book. The problem statement says:
Integrate \nabla \times{F},F=(3y,-xz,-yz^2) over the portion of the surface 2z=x^2+y^2 under the plane z=2, directly and using Stokes theorem.

So I started solving the integral directly. For the rotational I got:

\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3)

Then I parametrized the surface.:

\begin{matrix}x=x=r\cos \theta\\ y=r \sin \theta \\z=\displaystyle\frac{r^2}{2}\end{matrix}, \theta[0,2\pi] ,r[0,2]

Then I did: T_r\times{T_{\theta}}

T_r=(\cos \theta,\sin \theta,r),T_{\theta}=(-r\sin \theta,r\cos \theta,0)

For the cross product I got:
T_r\times{T_{\theta}}=(r^2\cos \theta,-r^2\sin \theta,r)

This is wrong. It should be \left<-r^2cos(\theta), -r^2sin(\theta), r\right> (oriented "upward").

And then I computed the integral

\displaystyle\int_{0}^{2}\int_{0}^{2\pi}(r\cos \theta-\displaystyle\frac{r^4}{4},0,\displaystyle\frac{-r^2}{2}-3)\cdot{(-r^2\cos \theta,-r^2\sin \theta,r)}d\theta dr=-12\pi

The result given by the book is: 20\pi.

I don't know what I did wrong, and is one of the first exercises that I've solved for the stokes theorem, so maybe I could get some advices and corrections from you :)

Thank you in advance. Bye.

 

[Telemachus]

Thank you very much HallsofIvy. How did you realize that the orientation was negative? I can identify a positive oriented normal in the Cartesian coordinates, but I don't know how to do it in cylindrical or spherical coordinates.