I got these two sphere (both are hollow, sorry i cant illustrate it by my drawings). The small sphere with radius $r$ and big sphere with radius $2r$. I just want to find the area of the shaded region. (i hope it clear that the region is made by the small sphere cut out the large sphere)
I got the answer which is $2 \pi r^{2}$ which i think is half the curved surface area of the hemisphere but i don’t know why, to me it should be the area of a circle with radius $r$ but i may be wrong. Please give me the hints but not full solution
Thanks for your help.
The triangle connecting the centers of two spheres and one intersection point is an isosceles triangle with side lengths $(r, 2r, 2r)$. The angle $\theta_0$ between the legs is given by
$$\sin\frac{\theta_0}2 = \frac{\frac{r}2}{2r} = \frac14 \implies \theta_0 = 2\arcsin \frac14$$
The surface you're after is a spherical cap with central half-angle $\theta_0$.
Its surface area is
$$A = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^{\theta_0} r^2\sin\theta \,d\theta \,d\phi = 2\pi r^2 \int_{0}^{\theta_0} \sin\theta\,d\theta = 2\pi r^2 (1-\cos\theta_0) = \frac14\pi r^2$$
since $$\cos\theta_0 = \sqrt{1- \sin^2\theta_0} = \sqrt{1-4\sin^2\frac{\theta_0}2\cos^2\frac{\theta_0}2} = \sqrt{1-4\sin^2\frac{\theta_0}2\left(1-\sin^2\frac{\theta_0}2\right)} = \frac78$$