Integrate area between circle and a function

[EudecioGabriel]

Hello!

I want to integrate the area between the function y(x)=x^2+y^2=1 and h(x)=|x|.

 

 

Can you help me?

 

Thank you!!

[EudecioGabriel]

Hello!

I want to integrate the area between the function y(x)=x^2+y^2=1 and h(x)=|x|.

 

 

Can you help me?

 

Thank you!!

[fiveworlds]

You need to find the points of intersection between the function and the circle.

 

In this case the point of intersection is sin(45) which is a recurring fraction.

Edited April 21, 2017 by fiveworlds

[imatfaal]

Similar threads merged

[Country Boy]

The line y= x intersects the unit circle, [math]x^2+ y^2= 1[/math] where [math]x^2+ x^2= 2x^2= 1[/math] or [math]x= \frac{\sqrt{\sqrt{2}}{2}[/math]

Similarly the line y= -x intersects the unit circle where [math]x= -\frac{\sqrt{2}}{2}[/math].

 

To find the area between y= |x| and the circle, integrate [math]\int_{-\sqrt{2}/2}^{\sqrt{2}/2} \sqrt{1- x^2}- |x| dx=[/math][math] 2\int_0^{\sqrt{2}/2} \sqrt{1- x^2}- x dx[/math].

Edited April 21, 2017 by Country Boy

[EudecioGabriel]

The line y= x intersects the unit circle, [math]x^2+ y^2= 1[/math] where [math]x^2+ x^2= 2x^2= 1[/math] or [math]x= \frac{\sqrt{\sqrt{2}}{2}[/math]

Similarly the line y= -x intersects the unit circle where [math]x= -\frac{\sqrt{2}}{2}[/math].

 

To find the area between y= |x| and the circle, integrate [math]\int_{-\sqrt{2}/2}^{\sqrt{2}/2} \sqrt{1- x^2}- |x| dx=[/math][math] 2\int_0^{\sqrt{2}/2} \sqrt{1- x^2}- x dx[/math].

COOL!! THANK YOU GUYS!

[Bender]

If you draw it, you see that it is one quarter of a circle

[Country Boy]

If you draw it, you see that it is one quarter of a circle

 

Well, if you want to do it the easy way. But that's no fun!

[Country Boy]

Actually, there are two bounded regions "between the circle x^2+ y^2= 1 and the graph y= |x|", one above the absolute value graph and below the circle, the other above the circle and below the absolute value graph.  As Bender said, the first area is a quarter the area of the entire circle, pi/4, the second is the other three quarters the area of the circle, 3\pi/4.

Edited August 2, 2017 by Country Boy