What's the surface area of the sphere $x^2 + y^2 + z^2 = 1$ over the disc $(x-1/2)^2 + y^2 \le 1/4$ ?
I've tried something, but I don't think it's right, as it's not a "nice answer"
So here is what I've done:
Firstly I parameterized using $x = r \cos(t)$, $y= r \sin(t)$, $z=z$.
$$dS = \sqrt{1+(dz/dx)^2+(dz/dy)^2} dA$$
$$dz/dx=2x , dz/dy=2y $$
Which gives me $$ \sqrt{1+4x^2+4y^2} dA = \sqrt{1+4r^2} r dr dt$$
Integral then becomes: S = $\int_0^{2\pi} \int_0^{1/2}\sqrt{1+4r^2} r dr dt$
Solving this gives me the result $=0,96$, however I don't think it's right
The border of the region in polar coordinates is: $$(r\cos\theta - 1/2)^2 + (r\sin\theta)^2 = 1/4,$$ $$r = \cos\theta,$$ with $\theta\in\cdots$ (Draw the circle to see the interval of variation of $\theta$).