Finding the area of the loop of the curve y^2=x^3(1-x)^2

[stardust006]

Find the area of the loop of the curve y^2=x^3(1-x)^2 using integral calculus.

y=√x^3(1-x)^2
y=√x^3/2 (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? >.<

 

[SammyS]

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(1-x)^2)    
y=√x^3/2 (1-x)    This should be y = ±√(x³) (1-x) or y = ±x^3/2 (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 ?

 

[stardust006]

Thank you, I should practice my math T.T