Problem: Consider the paraboloid $M$ parametrized by $X(u,v)=(u\cos v,u\sin v,u^2)$, $0\leq u, 0\leq v\leq 2\pi$. Denote by $M_r$ that portion of the paraboloid defined by $0\leq u\leq r$. Shifrin $3.1.8.$
Attempt: Since $\kappa_g=\alpha''\cdot (N\times\alpha')$, I let $x(v)=u\cos v;\quad\!\! y(v)=u\sin v; \quad\!\!z(v)=u^2=x^2+y^2$ and the curve is now parametrized as $\alpha(v)=(x(v),y(v),z(v))=(u\cos v,u\sin v, u^2)$z, which is a circle since $u$ is a constant. However, this leads to $z'(v)=z''(v)=0$ which implies $$\kappa_g=\alpha''\cdot(N\times \alpha')=0$$ I am not very familiar with differential geometry but I don't believe that the question would ask about $\displaystyle\int_{\partial M_r} \kappa_gds\quad\!\!$ if $\kappa_g$=0. So, how can I go about correctly parametrizing the curve?
Edit: $\alpha(v)=(u\cos v,u\sin v, u^2);\quad\!\!\!\!\alpha'(v)=(-u\sin v,u\cos v,0)\Rightarrow ||\alpha'(v)||=u$. Then $s(\alpha)=\displaystyle\int_0^t||\alpha'(v)||dv=ut$. So, with the arclength parametrization $$\alpha(s)=(u\cos(s/u),u\sin(s/u),u^2)$$ Then, $N=\alpha(s)/u$ and so $\kappa_g=\alpha''\cdot(N\times\alpha')=u^2(u\sin^2(u)+u\cos^2(u))=u^3$. I am not convinced that my approach was correct.
Sarah, that formula presumes an arclength parametrization. You can think about this geometrically (using right-angle trigonometry, like in Figure 3.1 of chapter 2), or you can use the chain rule (as often in the course), or you can reparametrize by arclength (see the bottom of p. 7 for reparametrizing a circle by arclength).
Nevertheless, I don't understand how you inferred from $z=r^2$ that that quantity would be $0$. You have a circle of radius $r$, so the vector $\kappa N$ points horizontally inward toward the center of the circle with magnitude $1/r$. Now what component of that vector is normal to the paraboloid and what component points in the direction of $\mathbf n\times\mathbf T$?