Find the area of the part of the sphere $S = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 = 1 \} $ cut by the $z \geq 59· \sqrt{x^2+y^2} $ cone.
I have tried to find the intersection curve but then I don't know what else to do. Should I change to spherical coordinates?
Any hint?
Thanks in advance.
On
In cylindrical coordinates, the two given surfaces are $r^2+z^2=1$ and $z=59r$. They intersect at the radius
$$r=\frac1{\sqrt{1+59^2}} $$
Given that the radius is fairly small, the spherical cutout is of a tiny circle and its area can then be approximated as
$$\pi r^2 \approx \frac{\pi}{59^2}$$
Hint: Actually, you are asked to evaluate the given integral $$\iint_{S} dS; \text {where, S is the surface of the given sphere cut by the given cone. }$$ Now, take the projection of the intersection curve on $xy$-plane, we get circle $x^2+y^2=\frac{1}{1+59^2}$ and we use the surface of the sphere to find $dS=\sqrt{1+z_x^{2}+z_y^{2}}dxdy=\frac{1}{|z|}dxdy=\frac{1}{\sqrt{1-x^2-y^2}}dxdy$. We use polar coordinate system to reduce the calculations, the integral becomes: $$\text{Surface area}= \int_{\theta=0}^{2\pi}\int_{r=0}^{\frac{1}{\sqrt{1+59^2}}} \frac{1}{\sqrt{1-r^2}} r dr d{\theta}.$$