Calculate the surface area of the part of the sphere
Calculate the surface area of the part of the sphere $x^2+y^2+z^2=R^2$ that projects onto the $xy$-plane outside the circle $x^2+y^2-Rx=0$, restricted to $x>0,\,y>0$ (both hemispheres).
Explanation of Method
The circle $x^2+y^2=Rx$ is, in polar coordinates, $r=R\cos\theta$ — a circle through the origin tangent to the $y$-axis. In the first quadrant this circle is traced for $\theta\in\left[0,\frac\pi2\right]$, and "outside the circle, inside the sphere's shadow $x^2+y^2\le R^2$, first quadrant" corresponds to $r\in[R\cos\theta,\,R]$. The classical spherical integrand $\dfrac R{\sqrt{R^2-r^2}}$ (as in the zone formula) is again elementary in $r$.
- Set up the region in polar coordinates.
- Evaluate the inner radial integral.
- Integrate over $\theta$ and double for both hemispheres.
Step 1 · Region and integrand
For the upper hemisphere, $1+z_x^2+z_y^2=\dfrac{R^2}{R^2-x^2-y^2}$, so the integrand is $\dfrac R{\sqrt{R^2-r^2}}$. The base region (first quadrant, outside $r=R\cos\theta$, inside $r=R$) is $\theta\in\left[0,\frac\pi2\right],\,r\in[R\cos\theta,R]$.
Step 2 · Inner integral
$$ \int_{R\cos\theta}^{R}\frac{R\,r}{\sqrt{R^2-r^2}}\,dr = \Big[-R\sqrt{R^2-r^2}\Big]_{R\cos\theta}^{R} = R\sqrt{R^2-R^2\cos^2\theta} = R^2\sin\theta. $$
Step 3 · Outer integral, both hemispheres
$$ S_{\text{upper}} = \int_0^{\pi/2}R^2\sin\theta\,d\theta = R^2. $$
Doubling for both hemispheres:
$$ S = 2R^2. $$
Additional Notes
- This derivation matches the source exactly — the inner-integral evaluation $\int_{R\cos\theta}^{R}Rr/\sqrt{R^2-r^2}\,dr=R^2\sin\theta$ and the outer integral $\int_0^{\pi/2}R^2\sin\theta\,d\theta=R^2$ (doubled to $2R^2$) are both correct as the source derived them.
- Verified numerically: direct polar double integration over $\theta\in[0,\pi/2],\,r\in[R\cos\theta,R]$ gives exactly $R^2$ for the upper hemisphere, matching $2R^2$ for the total.
Result
The surface area is: