Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$ Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$.
I drew the picture and found two surfaces: $S_1$ and $S_2$,
where $S_1$ is $\{ 0 \leq y \leq 1 \land -\sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \leq x \leq \sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \} \cap y + z = 1 \} $
and
$S_2$ is $\{ x^2 + y^2 + z^2 \leq 1 \cap x^2 + y^2 \leq y \} $ .
I wonder if this is correct:
$P(S_1) = \begin{gather*}
    \iint_D \sqrt{1+ 0^2 +(-1)^2}\,dx\,dy = ... = \frac{\sqrt{2}}{4}\pi
\end{gather*}
$
and
$P(S_2) = \begin{gather*}
    \iint_D \sqrt{1+ \frac{x^2}{1-x^2 - y^2} +  \frac{y^2}{1-x^2 - y^2}} \,dx\,dy = \cdots =\pi
\end{gather*}
$
In both cases $D$ is $\{ -\frac{1}{2} \leq x \leq  \frac{1}{2} $ and $ \frac{1}{2} - \sqrt{(\frac{1}{2})^2 - x^2}  \leq y \leq  \frac{1}{2} + \sqrt{(\frac{1}{2})^2 - x^2}\}$
Could someone please say if the idea is good, and if it isn't, what other approach should I take?
I am most familiar with this one.
Thanks in advance.
**EDIT **
There was a mistake in text above, I've corrected it.
 A: In cylindrical coordinates, the cylinder $x^2 + y^2 = y \ $ is
$r^2 = r \sin \theta \implies r = \sin \theta, \ 0 \leq \theta \leq \pi$.
This is a cylinder of radius $\frac{1}{2}$ with center at $(0,\frac{1}{2})$.
Now parametrization of cylinder is
$r(\theta, z) = (r\cos\theta, r\sin\theta, z) = (\frac{1}{2} \sin 2\theta, \sin^2\theta, z)$
$|r'_\theta \times r'_z| = 1$
Also note that at the intersection of the cylinder and the plane $y +z = 1$,
$\sin^2\theta + z = 1 \implies z = \cos^2\theta$
Also the intersection of the cylinder and the sphere $ \ \sin^2\theta + z^2 = 1 \implies z = \pm \cos\theta$
So the integral to find surface area of the cylindrical surface satisfying the given condition is
$\displaystyle 2 \int_{0}^{\pi/2} \int_{\cos^2\theta}^{\cos\theta} dz \ d\theta$
You may have to find for other surfaces similarly. The integral for other surfaces based on their parametrization are given by,
For spherical surface: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \frac{r}{\sqrt{1-r^2}} \ dr \ d\theta$
For planar bottom: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \sqrt2 \ r \ dr \ d\theta$
