# surface area with integrals

I'm working on a problem in my textbook and am confused on how to set up the integral.

"Find the surface area of the part of the hyperbolic paraboloid $$z= x^2 - y^2$$ that lies in the first octant and is inside the cylinder $$x^2 +y^2 = 1$$.

I have figured out the equation to integrate: $$\sqrt{4x^2+4y^2+1}$$ from finding the partial derivatives and taking the magnitude of the cross product.

You need to parameterize the integration region. One possible choice is to take $$\mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(u\cos v,u\sin v,u^2\cos2v)$$ with $$0\le u\le1$$ and $$0\le v\le2\pi$$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $$x^2+y^2=1$$. Then the $$z$$ coordinate is determined by the equation of the surface, $$z=x^2-y^2=u^2\cos^2v-u^2\sin^2v=u^2\cos2v$$
Then the surface element is $$\mathrm dS=\left\|\frac{\partial\mathbf s}{\partial u}\times\frac{\partial\mathbf s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$$ The area of the region of interest (call it $$R$$) is given by the surface integral, $$\iint_R\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv=\frac{(5\sqrt5-1)\pi}6$$