how to integrate the area between a circle and the x axis

[Oiler]

so far I have got to
[MATH](x-x_o)^2 +(y-y_o)^2=r^2[/MATH][MATH](y-y_o)^2=r^2 - (x-x_o)^2[/MATH][MATH]y-y_o=\sqrt{r^2-(x-x_o)^2}[/MATH][MATH]y=\sqrt{r^2 -(x-x_o)^2}+y_o[/MATH][MATH]\int_{x_1}^{x_2} \left(\sqrt{r^2 -(x-x_o)^2}+y_o\right)dx[/MATH]
but I cannot figure out how to solve for the anti derivative

[ wolfram misinterprets Mathjax, don’t know how to type it in a way it can understand, and symbolab spits out a monster ]

 

[Harry_the_cat]

Can you state your exact question as has been given to you?

 

[HallsofIvy]

Your integrand is $\displaystyle \sqrt{r^2- (x- x_0)^2}+ y_0$.
Of course $\displaystyle y_0$ is a constant and its integral is just $\displaystyle y_0(x_1- x_0)$.

So you are really concerned with $\displaystyle \int_{x_0}^{x_1}\sqrt{r^2- (x- x_0)^2}dx$.

The first thing I would do is make the substitution $\displaystyle z= x- x_0$. Then dx= dz, when $\displaystyle x= x_0$, $\displaystyle z= x_0- x_0= 0$ and when $\displaystyle x= x_1$, $\displaystyle z= x_1- x_0$ so the integral becomes
$\displaystyle \int_0^{x_1- x_0}\sqrt{r^2- z^2}dz$.

You should recognize that as leading to a "trig substitution". $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ so
$\displaystyle 1- sin^2(\theta)= cos^2(\theta)$ and $\displaystyle \sqrt{1- sin^2(\theta)}= cos(\theta)$.

Let $\displaystyle z= rsin(\theta)$. Then $\displaystyle dz= rcos(\theta)d\theta$ and $\displaystyle \sqrt{r^2- z^2}= \sqrt{r^2- r^2sin^2(\theta)}= r\sqrt{1- sin^2(\theta)}= r cos(\theta)$. When $\displaystyle z= 0$, $\displaystyle \theta= 0$ and when $\displaystyle z= x_1- x_0$, $\displaystyle \theta= aarcsin(x_1- x_0)$ so the integral is, finally,
$\displaystyle \int_0^{arcsin(x_1- x_0)} cos^2(\theta) d\theta$.

 

[Zermelo]

so far I have got to
[MATH](x-x_o)^2 +(y-y_o)^2=r^2[/MATH][MATH](y-y_o)^2=r^2 - (x-x_o)^2[/MATH][MATH]y-y_o=\sqrt{r^2-(x-x_o)^2}[/MATH][MATH]y=\sqrt{r^2 -(x-x_o)^2}+y_o[/MATH][MATH]\int_{x_1}^{x_2} \left(\sqrt{r^2 -(x-x_o)^2}+y_o\right)dx[/MATH]
but I cannot figure out how to solve for the anti derivative

[ wolfram misinterprets Mathjax, don’t know how to type it in a way it can understand, and symbolab spits out a monster ]

Have you ever worked with parametric functions and integrals? The parametric equation for the circle is just x = cos t, y = sin t. Now you need to solve Integral f(x) dx, dx = x'(t) dt, so you need to solve Integral sint (cos'(t)) dt, which is much easier