IntegrateMe Messages 214 Reaction score 1 Thread starter Mar 18, 2012 #1 Why is the area of the cardioid r = 2sinθ-2 in the first quadrant represented with limits from π to 3π/2 and not with limits from 0 to π/2? Isn't the first quadrant 0 to π/2?
Why is the area of the cardioid r = 2sinθ-2 in the first quadrant represented with limits from π to 3π/2 and not with limits from 0 to π/2? Isn't the first quadrant 0 to π/2?
awkward Messages 364 Reaction score 0 Mar 18, 2012 #2 r is negative (or zero) for all values of theta, so when theta is between pi and 3 pi / 2, (r, theta) is in the first quadrant.
r is negative (or zero) for all values of theta, so when theta is between pi and 3 pi / 2, (r, theta) is in the first quadrant.
IntegrateMe Messages 214 Reaction score 1 Mar 18, 2012 #3 Actually, never mind, I'm still not understanding this :( Last edited: Mar 18, 2012
awkward Messages 364 Reaction score 0 Mar 18, 2012 #4 Try plotting the point for theta = 5 pi / 4. Maybe you will see it then.