Find the surface area of the part of the paraboloide $z=\frac{x^2+y^2}{2}$ inside the sphere $x^2+y^2+z^2=3$

Setting $2z = x^2+y^2$ I obtain that the points of intersection between the paraboloide and the sphere are $2z+z^2=3$ and thus $z=1$ is the point of intersection, since $z=-3$ is not a solution.

Now in the solution, they parametrize the surface $S$ to $$\vec r(r, \theta) = [rcos(\theta), rsin(\theta), \frac{1}{2}r^2]$$

Since $z=1$ from $2z=r^2$ we obtain that $0 \leq r \leq \sqrt{2}$

$$ dS = |\frac{\partial \vec r}{\partial r} \times \frac{\partial \vec r}{\partial \theta}| = r\sqrt{r^2+1} drd\theta$$ and thus the surface area is given by

$$ S = \int_0^{2\pi} \int_0^{\sqrt{2}} dS drd\theta$$

My main problem here is understanding the parametrization $$\vec r(r, \theta) = [rcos(\theta), rsin(\theta), \frac{1}{2}r^2]$$How does this parametrize the surface we're interested in? I can understand the components, as we're looking for the surface area of the paraboloide INSIDE the sphere. But the parabloide has a "roof", it gets "cut" by the sphere at $z=1$. How does the limit $0 \leq r \leq \sqrt{2}$ account for that? It seems to me they're just integrating some integrand over a circle (the projection of the curve of intersection) without accounting for the "height" here. Can someone shed some light on this for me, please?

Thank you in advance


1 Answer 1


The parametrization $\vec r(r, \theta) = [r\cos(\theta), r\sin(\theta), \frac{1}{2}r^2]$ works since

  • $x=r\cos(\theta)$
  • $y=r\sin(\theta)$
  • $z= \frac{x^2+y^2}{2}=\frac{1}{2}r^2$

is such that for $\theta\in[0,2\pi]$ and $r\in[0,\sqrt 2]$ we can represent each point $(x,y,\frac{x^2+y^2}{2})$ of the paraboloid cuted by the sphere.

Note that the limit for $r_{max}=\sqrt 2$ comes from the condition that for $z=1$ we have

$$z= \frac{x^2+y^2}{2}=1\implies r^2=x^2+y^2=2 \implies r=\sqrt 2$$


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