Finding the surface area $\iint_{s} f \, dS$ of $z=x^2-y^2$ cut off by $z=4-2y^2$

Question

Finding the surface area $\iint_{s} f \, dS$ of $z=x^2-y^2$ cut off by $z=4-2y^2$

I have no idea which parametrization to use for this, however i did figure out the following:

I think the points where the two curves intersect can be used as limits for something of which are at: $$x^2-y^2=4-2y^2$$ $$x^2+y^2=4 \tag{1}$$

I think this means that the two curves form a circle where they intersect, does this mean that i only need to integrate the region at $(1)$ only?.

I also computed this: $$\sqrt{ \left( \frac{\partial z}{\partial r} \right)^2 + \left( \frac{\partial z}{\partial \theta} \right)^2 + 1 } = \sqrt{4(x^2+y^2)+1}$$

I noticed the two occurences of the $x^2+y^2$ of which can be changed into $r^2$, assuming that i am only integrating the region in $(1)$ i would have:

$$\iint_{s} f \, dS=\int_0^{2\pi}{} \int_0^2{\sqrt{4r^2+1}dr d\theta}$$

Is my assumption correct?, please help me with the parametrisation -Thanks.

Answer 1

I would say, tasks properly, only when it is necessary to install the coordinate transformation Jacobian:

$\displaystyle \iint_{s} f \, dS=\int_0^{2\pi}{} \int_0^2{\sqrt{4r^2+1}\,|J|\,dr d\theta}$

$\displaystyle J=\begin{vmatrix} x'_r &x'_\theta \\ y'_r & y'_\theta \end{vmatrix}=\begin{vmatrix} \cos \theta & -r\sin \theta \\ \sin \theta & r\cos\theta \end{vmatrix}=r$

so

$\displaystyle \iint_{s} f \, dS=\int_0^{2\pi}{} \int_0^2{r\sqrt{4r^2+1}\,dr d\theta}=2\pi\int_{0}^{2}r\sqrt{4r^2+1}\,dr$

A last integral substitution $4r^2+1=t^2,\quad r\,dr=t/4\,dt$