EudecioGabriel Posted April 21, 2017 Share Posted April 21, 2017 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!! Link to comment Share on other sites More sharing options...
EudecioGabriel Posted April 21, 2017 Author Share Posted April 21, 2017 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!! Link to comment Share on other sites More sharing options...
fiveworlds Posted April 21, 2017 Share Posted April 21, 2017 (edited) 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 1 Link to comment Share on other sites More sharing options...
imatfaal Posted April 21, 2017 Share Posted April 21, 2017 Similar threads merged 1 Link to comment Share on other sites More sharing options...
Country Boy Posted April 21, 2017 Share Posted April 21, 2017 (edited) 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 1 Link to comment Share on other sites More sharing options...
EudecioGabriel Posted April 21, 2017 Author Share Posted April 21, 2017 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! Link to comment Share on other sites More sharing options...
Bender Posted April 24, 2017 Share Posted April 24, 2017 If you draw it, you see that it is one quarter of a circle 1 Link to comment Share on other sites More sharing options...
Country Boy Posted June 23, 2017 Share Posted June 23, 2017 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! Link to comment Share on other sites More sharing options...
Country Boy Posted August 2, 2017 Share Posted August 2, 2017 (edited) 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 Link to comment Share on other sites More sharing options...
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