Find the area inside the lemniscate ,which is described by the equation (x^2 + y^2) = 2a^2 (x^2 - y^2) I have no idea where to start so I found out what a lemniscate is.. but this didn't help me much. I have rearranged the equation for a: a = sqrt ( (x^2 + y^2)^2 / 2(x^2 - y^2) ) a = x^2 + y^2 / sqrt (2(x^2 - y^2)) How do I proceed from here?
what does the a even mean.. my prof did not say anything about a lemniscate, I am confused out of my mind
okay so i can only put it in polar coordinates, but where are the limits of integration in which i am suppose to integrate?? (r^2)^2 = 2a^2 (rcos^2theta - rsin^2theta) r r^4 = 2a^2 (rcos^2theta - rsin^2theta) r r^3 = 2a^2 (r^2cos^2theta - r^2sin^2theta) r = 2a^2 (cos^2theta - sin^2theta) r = 2a^2 (1-sin^2theta - sin^2theta) r = 2a^2 (1-2sin^2theta) Using half angle rule: r = 2a^2 (1-2(1-cos^2theta / 2)) r = 2a^2 (1-(1-cos^2theta)) r = 2a^2 (cos^2theta) i still do not know the meaning of a.. and the boundaries
should be r^2 = 2a^2 (cos^2theta); sketch the curve to find the limits of intergration, I'd say for the right loop you integrate from -pi/4 to pi/4. Constant a controls the form of the curve. Try sketching it here http://graph.seriesmathstudy.com/
Well, you don't. What is the integral of y = ax ? ax^2 / 2, right? Can you integrate r^2 = 2a^2 (cos^2theta) then, integrate wrt to theta not a?
i was doing the same problem and i got 2a^2 as my answer r^2 = 2a^2cos(2theta) the integral of r^2 should be the area hence integrating the right side gives u an asnwer wrt a
anyone know how to compute Integral 0 to 1 ; integral y to 1 (1+x^2)^(1/2) dxdy its a double integral problem, im stuck....
i found examples in class where it is integrating the double integral w.r.t dr dtheta so do i rearrange for a to get rid of a... or do as u say and integrate w.r.t dtheta i still have not found the limits of integration yet, i dont see how i am suppose to figure all this stuff out with no information and no knowledge of a lemniscate
i still need help on this, anyone have any ideas? is hanhan correct that r^2 is the area so u can integrate ??