How do I find the Integral of $\sqrt{r^2-x^2}$? How can I find the integral of the following function using polar coordinates ?

$$f(x)=\sqrt{r^2-x^2}$$ 

Thanks!
 A: since  $y=r\sin { \theta  } .\\ x=r\cos { \theta  } $

$$\\ \int { \sqrt { r^{ 2 }-x^{ 2 } }  } dx=-\int { \sqrt { { r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta  }  }  } r\sin { \theta  } d\theta =-{ r }^{ 2 }\int { \left| \sin { \theta  }  \right| \sin { \theta d\theta  } =\pm \frac { { r }^{ 2 } }{ 2 } \left( \theta -\frac { \sin { 2\theta  }  }{ 2 }  \right)  } +C$$

A: It would help to know what you know, so that I can give an informed answer. Since I don't, I will try to guess your level. Also, it would help if you would specify what you want to integrate over. Are you trying to calculate an indefinite integral or a definite one over all of $\mathbb{R}$?
A definitive integral over all $\mathbb{R}$ in polar coordinates takes the following form:
$$\int_{x=-\infty}^{\infty} f(x)dx = -\int_{r=0}^{\infty} \int_{\theta= 0}^{2\pi}f(r,\theta)r sin(\theta) d\theta dr.$$
Your function is $f (r, x) = \sqrt{r^2-x^2}$ but you need $f(r, \theta)$, not $f(r, x)$ to be able to integrate in polar coordinates. The common change from cartesian to polar coordinates goes as follows:
$$x= r \cos\theta, \\y =r\sin\theta.$$
If you are unfamiliar with what I've said so far, check this: http://mathworld.wolfram.com/PolarCoordinates.html
So far so good, we need to move from $f(r, x)$ to $f(r, \theta)$ and we can do this via $f(r, \theta) = f(r, x(\theta))$, i.e. we plug in the definition of the coordinate change $x = r cos \theta$ and obtain:
$$f(r, \theta) = \sqrt{r^2-(r \cos\theta)^2}= \sqrt{r^2 - r^2 \cos^2(\theta)} = \sqrt{r^2(1-\cos^2(\theta))} = r \sqrt{\sin^2(\theta)}$$
Can you take it from here and do the integral above now?
A: Using the substitution $x=r\cos{\theta}$    we have $dx = -r\sin{\theta}\;d{\theta},$ hence
$$\int\sqrt{r^2-x^2}\;dx = -\int\sqrt{r^2-r^2\cos^2{\theta}}\cdot r\sin{\theta}\;d{\theta} = \\
= -r^2\int |\sin{\theta}|\cdot \sin{\theta}\;d{\theta}=
\begin{cases}
-r^2\int \sin^2{\theta}\;d{\theta} \ \ \text{if} \ \ \sin{\theta}>0, \\
r^2\int \sin^2{\theta}\;d{\theta} \ \ \text{if} \ \ \sin{\theta}<0.
\end{cases}$$
Then reduce degree by the identity $\sin^2{\theta}=\frac{1-\cos{2{\theta}}}{2} $    and integrate it.
A: $\displaystyle\int \sqrt{r^2-x^2}dx$
Let be $\;x=r.\sin\alpha$   or   $\quad x=r.\cos\alpha$,
Let be $\;x=r.\sin\alpha$,
and $\quad dx=r.\cos \alpha \;d\alpha$
Integral be,
$\displaystyle\int \sqrt{r^2-x^2}dx=\displaystyle\int r.\sqrt{1-\sin^2\alpha}\;.r.\cos \alpha \;d\alpha=\displaystyle\int r^2.\cos^2\alpha\; d\alpha$
And use this equation,$\quad \cos2\alpha=2\cos^2\alpha-1\longrightarrow \cos^2\alpha=\dfrac{\cos 2\alpha+1}{2}$
Then,
$=\displaystyle\int r^2.\cos^2\alpha\; d\alpha=\displaystyle\int r^2.\dfrac{\cos 2\alpha+1}{2}\; d\alpha=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C$
From here ,
$\boxed{\;x=r.\sin\alpha}$
Answer'll be,
$\boxed{\boxed{\displaystyle\int \sqrt{r^2-x^2}dx=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C=\dfrac{r^2}{2}\left[\dfrac{2x\sqrt{r^2-x^2}}{r^2}+\arcsin\left(\dfrac{x}{r}\right)\right]}}$
