Step 1: Identify the curve:
Squaring $y=2\sqrt{1-x^2}$ gives $y^2=4(1-x^2)$, so $4x^2+y^2=4$, that is $\frac{x^2}{1}+\frac{y^2}{4}=1$. This is an ellipse with $a=1$, $b=2$. Since $y\geq0$ we have the upper half.
Step 2: Use the ellipse area:
The whole ellipse has area $\pi ab=\pi(1)(2)=2\pi$.
Step 3: Take half:
The required region is the upper half, so the area is $\frac{2\pi}{2}=\pi$.
Step 4: Cross-check by substitution:
Put $x=\sin\theta$. Then $\int_{-1}^{1}2\sqrt{1-x^2}dx=2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=2\cdot\frac{\pi}{2}=\pi$.
Final Answer:
The area is $\pi$ (option C).
\[ \boxed{\pi} \]