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!!

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!!

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, 20178 yr by fiveworlds

Similar threads merged

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, 20178 yr by 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].

COOL!! THANK YOU GUYS!

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

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!

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, 20178 yr by Country Boy