Find the area below $y=\sqrt{4-x^2}$ and above the x-axis between $x=0$ and $x=2$ I'm trying to figure out how to go about this, but I'm not entirely sure. Is there a specific way to find the area under a curve or line (and strictly above the x-axis) before doing Riemann sums?
 A: It is quarter of the circle whose radius is $2$ and center origin, so the answer is $\pi$.

You can also substitute $x=2\sin \theta\;(0 \le \theta \le \frac{\pi}{2})$, then
\begin{align}
\int_0^2 \sqrt{4-x^2}dx &= \int_0^{\frac{\pi}{2}} 2\cos \theta\sqrt{4-4\sin^2\theta}  d\theta\\
&=\int_0^{\frac{\pi}{2}} 4\cos^2\theta d\theta\\
&=\int_0^{\frac{\pi}{2}} 4 \frac{1+\cos 2\theta}{2}d\theta\\
&=\left[2\left(\theta+\frac{1}{2}\sin2\theta\right)\right]_0^{\frac{\pi}{2}}\\
&=\pi
\end{align}
A: substitute $$x=2\sin { \theta  } \\ dx=2\cos { \theta d\theta  } $$ and change bounders so that:
$$\int _{ 0 }^{ 2 }{ \sqrt { 4-{ x }^{ 2 } } dx } =\int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \sqrt { 4-4\sin ^{ 2 }{ \theta  }  }  } 2\cos { \theta d\theta  } =4\int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \cos ^{ 2 }{ \theta  } d\theta = } 2\int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \left( 1+\cos { 2\theta  }  \right) d\theta = } 2{ \left( \theta +\frac { \sin { 2\theta  }  }{ 2 }  \right)  }_{ 0 }^{ \frac { \pi  }{ 2 }  }=\pi \\ \\ $$
