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Feb28-12, 09:19 AM
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Quote Quote by stardust006 View Post
Find the area of the loop of the curve y^2=x^3(1-x)^2 using integral calculus.
                   The following needs extra parentheses to be correct.
y=√x^3/2 (1-x)    This should be y = √(x3) (1-x) or y = x3/2 (1-x) .

To sketch the curve, I assigned values for x and then solved the corresponding values of y.

x= -1, y= -2
x= -0.5, y= -0.53
x=0, y= 0
x= 0.5, y= 0.177
x=1, y=0

how can i find the area of this? >.<
Hello stardust006. Welcome to PF !

The is important. It gives a clue as to why the graph has a loop.

There are two x-intercepts. Can you find them ?