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

O

Oiler

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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 ]
 
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,
\(\displaystyle \int_0^{arcsin(x_1- x_0)} cos^2(\theta) d\theta\).
 
Oiler said:
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 ]
Click to expand...

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 :)
 
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