I have the following homework problem:

In part, we compute both integrals from Stokes' theorem for the helicoid $\displaystyle {\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)}$ where $\displaystyle {(r,\theta)}$ lies in the rectangle $\displaystyle {[0,1] \times [0,\pi/2]}$, and $\displaystyle {\mathbf{F}}$ is the vector field $\displaystyle {\mathbf{F} = (5 z, 3 x, 7 y)}$.

First, compute the surface integral:

$\displaystyle {\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta}$, where

$\displaystyle {a = }$ _, $\displaystyle {b = }$ _, $\displaystyle {c = }$ _, $\displaystyle {d =}$ _, and

$\displaystyle {f(r, \theta) = }$ ____ (use "t" for theta).

Finally, the value of the surface integral is ____.

Next compute the line integral on that part of the boundary from $\displaystyle {(1,0,0)}$ to $\displaystyle {(0,1,\pi/2)}$.

$\displaystyle {\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}$, where

$\displaystyle {a =}$ _, $\displaystyle {b =}$ _,

and $\displaystyle {g(\theta) =}$ ____ (use "t" for theta).

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So far I have a=0, b=pi/2, c=0, d=1, and those are right. But, I'm stuck at $\displaystyle {f(r, \theta) = }$ ____. I tried finding:

$\displaystyle {\Psi_r = (\cos \theta, \sin \theta, 0)}$

$\displaystyle {\Psi_\theta = (-r\sin \theta, r\cos \theta, 1)}$

$\displaystyle {\Psi_r \times \Psi_\theta = (\sin \theta, - \cos \theta, r)}$

Then I plugged $\displaystyle {\Psi}$ into $\displaystyle {\mathbf{F} = (5 z, 3 x, 7 y)}$ to get:

$\displaystyle (5 \theta, 3r \cos \theta, 7r \sin \theta) $

I dotted that with the cross product and found:

$\displaystyle 5 \theta \sin \theta - 3r \cos^2 \theta + 7 r^2 \sin \theta $

Which is what I thought $\displaystyle {f(r, \theta) = }$ should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?