Deciding the limits of $\theta$ for this Stokes' Theorem problem 
Let S be the bounded surface of $z = x^2 + y^2$ intercepted by $z = ax + by$, where $a$ and $b$ are any two real numbers. Let $C$ be the bounding curve of $S$ traversed counterclockwise as viewed from the positive $z$ axis. Make a rough sketch showing the surface $S$, the curve $C$ and outward pointing unit normal vector to the surface $S$. Use Stokes' theorem to evaluate the integral $ \oint_C y^2 dz $

Parametrization of surface is: $$r(\theta, a) = t\cos\theta \hat i + t\sin \theta \hat j + t^2 \hat k $$ where $0 < t  <\sqrt{(a\cos \theta +b \sin \theta)} $
Curl = $ 2t \sin \theta \hat i $
$(\nabla \times F) \cdot (r_ t \times r_\theta ) = -4t^3 \cos \theta \sin \theta$
So by Stokes' Theorem the line integral can be evaluated as $\displaystyle \int \int_0 ^{(\sqrt{a^2/4 + b^2/ 4})}-4t^3 \cos \theta \sin \theta dt d\theta $
But I can't decide the limits of $\theta$. Seeing the inersection of plane and $xy$ plane, the limits in my opinion should be from $-\pi - \arctan(\frac ab)$ to $  -\arctan(-\frac ab)$. But with these limits, evaluating the double integral gives $0$. So could someone help with the correct limits.
 A: Hint: It is better if you consider another cobordant surface, you can try a disc in the plane. Thus, the disc surface has the parametrization
$$x=\frac{a}{2}+r\cos\theta,\quad y=\frac{b}{2}+r\sin\theta,\quad z=\frac{a^2+b^2}{2}+r(a\cos\theta+b\sin\theta),$$
where $0\leq\theta\leq2\pi$ and $0\leq r\leq\dfrac{a^2+b^2}{4}$.
On the other hand, you can show that curve C is always a circle of radius $(a^2+b^2)/4$ unless $a=b=0$. The center is $(a/2,b/2)$.
A: The paraboloid $z=x^2+y^2$ and the plane $z=ax+by$ intersect for
$$x^2+y^2=ax+by\implies\left(x-\frac a2\right)^2+\left(y-\frac b2\right)^2=\frac{a^2+b^2}4$$
i.e. in the cylinder with cross sections parallel to the $x$-$y$ plane centered at $\left(\frac a2,\frac b2\right)$ with radius $\frac{\sqrt{a^2+b^2}}2$. Naturally, I think this suggests a parametrization of $S$ in cylindrical coordinates,
$$\hat s(u,v)=x(u,v)\,\hat\imath+y(u,v)\,\hat\jmath+z(u,v)\,\hat k$$
where
$$\begin{cases}x(u,v)=u\cos v+\frac a2\\[1ex]y(u,v)=u\sin v+\frac b2\\[1ex]z(u,v)=x(u,v)^2+y(u,v)^2=\frac{a^2+b^2}4+u^2+au\cos v+bu\sin v\end{cases}$$
with $0\le u\le\frac{\sqrt{a^2+b^2}}2$ and $0\le v\le2\pi$.
From the integrand we can obtain the underlying vector field $\hat F$:
$$y^2\,\mathrm dz=\underbrace{(0\,\hat\imath+0\,\hat\jmath+y^2\,\hat k)}_{\hat F(x,y,z)}\cdot(\mathrm dx\,\hat\imath+\mathrm dy\,\hat\jmath+\mathrm dz\,\hat k)$$
Compute the curl:
$$\nabla\times\hat F=2y\,\hat\imath$$
Take the normal vector to $S$ to be $\hat n=\frac{\partial\hat s}{\partial v}\times\frac{\partial\hat s}{\partial u}$:
$$\hat n=\frac{\partial\hat s}{\partial v}\times\frac{\partial\hat s}{\partial u}=(au+2u^2\cos v)\,\hat\imath+(bu+2u^2\sin v)\,\hat\jmath-u\,\hat k$$
Then by Stokes' theorem the integral is
$$\begin{align*}
\oint_Cy^2\,\mathrm dz&=\oint_C\hat F\cdot\mathrm d\hat r\\[1ex]
&=\iint_S(\nabla\times\hat F)\cdot\mathrm d\hat S\\[1ex]
&=\int_0^{2\pi}\int_0^{\sqrt{a^2+b^2}2}\left(2y(u,v)\,\hat\imath\right)\cdot\hat n\,\mathrm du\,\mathrm dv\\[1ex]
&=2\int_0^{2\pi}\int_0^{\sqrt{a^2+b^2}/2}\left(u\sin v+\frac b2\right)\,\mathrm du\,\mathrm dv\\[1ex]
&=\boxed{2\pi b\sqrt{a^2+b^2}}
\end{align*}$$
