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 July 24th, 2013, 08:11 PM #1 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 surface integration Find the value of the area bounded by the curve x^4 +y^4 =x^2 +y^2.
 July 24th, 2013, 08:38 PM #2 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: surface integration Here is the graph of the function, so you know what I am talking about: https://www.desmos.com/calculator/pod5cz9ytj The graph of your function is the same as the combination of the blue curve and the red curve. Find the maximum and minimum x values of the red curve (which have been indicated in purple and green, respectively), then integrate the red curve with those values as the limits. Next, integrate the blue curves separately. The leftmost one will have limits at the green line and at -1, and the rightmost one will have limits at the purple line and at 1. Subtract the area under the red curve by the total area under the blue curve, and that will be your answers.
 July 26th, 2013, 03:30 AM #3 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 Re: surface integration Your solution is very interesting, but unfortunately I can't open the graph because my phone doesn't support. So can you give me the answer? I found it already so I just want to verify that I was right or wrong.
 July 26th, 2013, 07:33 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: surface integration If you want to verify your answer then tell us what you got.
 July 27th, 2013, 02:06 AM #5 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 Re: surface integration I got the value of area is equal to pi. x^4 +y^4 =x^2 +y^2, then we have (x^2 -1)x^2 +(y^2-1)y^2 =0. So I took x^2 -1=0, y^2 -1=0. I got x^2 +y^2 =2, and this is the circle equation. Let x=sqrt(2)cost, y=sqrt(2)sint. So I can find the area by using integral.

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