<s>[INST] Area of the intersection of a cone and a sphere Calculate the area of the surface $x^2+y^2 = z^2 $ with $z \geq 0$, limited by $x^2 +y^2+z^2=2ax$.
I think the method to solve it is to calculate the parametric equation of the curve and then calculate the area of the enclosed curve using $ \frac{1}{2} \int_c x\,dy - y\,dx $. I'm having trouble finding the equation of the parametric curve.
Thanks.
 [/INST] Easy way:
The projection of the "intersection" on the x-y plane is a circle of radius $\frac{a}{2}$.
Since the normals on the cone is making an $45^\circ$ angle with the unit z-vector,
the area of the intersection is $\frac{\pi}{4} a^2 / \cos{45^\circ} = \frac{\pi}{2 \sqrt{2}} a^2$.
Harder? way:
Introduce parametrization $(r,\theta) \rightarrow 
(\frac{r}{\sqrt{2}} \cos{\theta}, \frac{r}{\sqrt{2}} \sin{\theta}, \frac{r}{\sqrt{2}} )$
to the cone surface.
The cone and sphere intersect at:
$$
    r^2 = x^2 + y^2 + z^2 = 2ax = \sqrt{2} a r \cos{\theta}
\implies -\pi/2 \le \theta \le \pi/2  \text{ and } 0 \le r \le \sqrt{2} a \cos{\theta}.
$$
In terms of $(r,\theta)$, the metric is
$dx^2 + dy^2 + dz^2 = dr^2 + \frac{r^2}{2} d\theta^2$, 
the surface area element is $\sqrt{\frac{r^2}{2}} dr d\theta$ 
and the area we want becomes:
$$
   \frac{1}{\sqrt{2}} \int_{-\pi/2}^{\pi/2} d\theta \int_0^{\sqrt{2} a \cos{\theta}} r dr
=  \frac{a^2}{\sqrt{2}} \int_{-\pi/2}^{\pi/2}  \cos{\theta}^2 d\theta
=  \frac{\pi}{2 \sqrt{2}} a^2
$$
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