I was following the steps outlined in my professor's notes to solve this stokes theorem problem. In the end, I have an integral that not even online integral calculators can solve. The correct answer is $9\pi$
Let S be the part of the sphere $x^2+y^2+z^2=25$ which lies above the plane $z=4$. Orient S by upward pointing normal vectors. Evaluate. $$\int\!\!\!\int_{S}\text{curl}F\,dS$$ where $F=yze^xi+(x+ze^x)j+yzk$. (Hint: Consider a different surface)
I want to use stokes theorem for this question. It seems to be appropriate. It's also the only way I was taught where you can consider a different surface.
$x^2+y^2+z^2=25;z=4$. Formula for sphere chopped @ z=4: $x^2+y^2=9$; $r=3$ Parametrization: $\langle3\cos t,3\sin t,4\rangle$; derivative of parametrization: $\langle-3\sin t,3\cos t,0\rangle$
This is where things get messy. I need to plug my parameters into the vector field F and Dot it with deriv. of Parametrization. $$F=\langle(12\sin(t)e^{3\cos(t)}i,3\cos(t)+4e^{3\cos(t)}j,12\sin(t)k\rangle\cdot\langle-3\sin t,3\cos t,0\rangle$$ This equals: $$-36\sin^2(t)e^{3\cos(t)}+9\cos^2(t)+12\cos(t)e^{3\cos(t)}$$ Which I'm supposed to take the integral of and multiply it by the area of my disk, since i'm using the flat disk as my surface instead of the top of the sphere.
Evidently I must have made some kind of error, but I can't find it! The correct answer is $9\pi$.
Thanks in advance.
You missed the point, by not heeding the hint. The use of Stokes here is, as the hint says, that you can use any surface $S$ with the same boundary curve. For instance, you can use the disk $x^2+y^2\leq 3$, $z=4$. Although $F$ looks scary, its curl is not that bad: $$ \text{curl}\, F=\langle z-e^x,ye^x,1\rangle. $$ So, parametrizing $S'$ as $u(r,t)=\langle r\cos t,r\sin t,4\rangle$, we have $$ \frac{\partial u}{\partial r}\times\frac{\partial u}{\partial t}=\langle 0,0,r\rangle. $$ Then \begin{align} \int\!\!\!\!\int_S\text{curl}\,F\,dS &= \int\!\!\!\!\int_{S'}\text{curl}\,F\,dS =\int_0^3\!\!\!\int_0^{2\pi}\langle z-e^x,ye^x,1\rangle\cdot\langle 0,0,r\rangle\,dt\,dr\\ &=\int_0^3\!\!\!\int_0^{2\pi}\,r\,dt\,dr=2\pi\,\int_0^3r\,dr=2\pi\times\frac92=9\pi. \end{align}