Problem:
Compute the surface area of that portion of the sphere $x² + y² + z² = a²$ lying within the cylinder $x² + y² = ay$ , where a > 0.
My Try:
We consider $r(x,y)=(x,y,f(x,y))=(x,y,\sqrt{a²-x²-y²})$ so $$ ||\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y} ||=\sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2}=\sqrt{1+\frac{x²}{a²-x²-y²}+\frac{y²}{a²-x²-y²}}=\frac{a}{\sqrt{a²-x²-y²}}$$ and for the region (gonna use polar coordinates)then so $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we have $x²+y²\leq ay\rightarrow r²\leq ar\sin(\theta)\rightarrow r\leq a\sin(\theta)$ so $0\leq r\leq a\sin(\theta)$ and $theta \in [0,\pi]$ because $a>0$ then the circumference is in the positive $y$-axis.
$$ \iint_{R}^{}\sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2} dA =\iint_{R}^{}\frac{a}{a²-x²-y²} dA = \int_{0}^{\pi} \int_{0}^{a\sin(\theta)} \frac{ar}{\sqrt{a²-r²}} dr d\theta$$ $$= -a\int_{0}^{\pi}\int_{0}^{a\sin(\theta)} (\sqrt{a²-r²})' drd\theta=-a\int_{0}^{\pi} a(|\cos(\theta)|-1) d\theta= -a²\int_{0}^{\pi}|\cos(\theta)|-1d\theta=-a²(2-\pi)=(\pi-2)a²$$
where the last $\int_{0}^{\pi}|cos(t)|-1 dt=\int_{0}^{\frac{\pi}{2}}cos(t)dt+\int_{\frac{\pi}{2}}^{\pi}-\cos(t)dt-\pi= 1-(-1)-\pi=2-\pi $
The Book have answer and the answer of exercise is $(2\pi-4)a²$ but I got $(\pi-2)a²$, So Have I made a mistake in the resolution?
Hint:
Your work is correct, but there are two surfaces on the sphere with the area that you have calculated, one with $z=\sqrt{a^2-x^2-y^2}$ (above the $xy$ plane) and the other with $z=-\sqrt{a^2-x^2-y^2}$ (below the plane). See the figure.