find the area enclosed by the curve (parametric equation)by vande060 Tags: curve, enclosed, equation, parametric 

#1
Apr1911, 03:20 PM

P: 187

1. The problem statement, all variables and given/known data
I dont really have a problem with integration here, I just need to learn how to decide what direction the integration should be in find the area enclosed by the curve x = t^{2}  2t y = t^{1/2} around the y axis 2. Relevant equations A = ∫ xdy 3. The attempt at a solution so when i plug values into the parametric equations i find that this graph comes out of the origin at y = 0 then crosses the axis again at 2^{1/2} I feel like my bound of integration should be from 0 to 2^{1/2}, but I get a negative area, so I probably should reverse the bounds, but I cant rationalize the reversal. A = from 0 to 2^{1/2} ∫ (t^{2}  2t)t^{1/2} dt = from 0 to 2^{1/2} ( 4/5 *t^{5/2 }  4/3 * t^{3/2}) here i get 1.86  2.2, which is negative 



#2
Apr1911, 04:17 PM

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P: 7,423

Have you looked at the graph?




#3
Apr1911, 04:40 PM

P: 187

sure it looks like this approx
starting at t=0 there the curve is at the origin, and at point t = 2 the graph it at the point 0,2^{1/2} so the bounds i used seemed right 



#4
Apr1911, 05:59 PM

P: 242

find the area enclosed by the curve (parametric equation)
the area should be negative; it is 'below' the yaxis.
there is one mistake that i can see. if y = t^(1/2), then dy = (1/2)*t^(1/2) dt. you seem to have integrated xy instead of xdy. 



#5
Apr1911, 06:50 PM

P: 187





#6
Apr1911, 07:03 PM

P: 242

cheers



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