Calculus help (integration)?

Question

find the are bounded by y^2=x^2(1-x^2)
(its a graph like a sideways figure 8 with zeros at the origin and y=-1,+1)

Hint: find the region bounded by y=x(sqrt(1-x^2)) and y=0 (the area of 1 quarter of the figure 8)and multiply by four

this is what i did:
∫x(sqrt(1-x^2))dx
u=1-x^2
du=-2xdx
du/-2=xdx

i got it down to -1/2(2/3u^(3/2))

then when i eval it from 0 to 1 i get 1/3 and its supposed to be 4/3

can you tell me what i did wrong?

thanks!

oh haha i wrote all that sh** out for no reason! thanks!

Answer 1

∫x(sqrt(1-x^2))dx
let u = (1 - x^2)
du = -2x dx
xdx = -du/2
-(1/2)∫u^(1/2)du =
-(2/2*3)u^(3/2) =
-(1/3)(1 - x^2)^(3/2) from 0 to 1 =
-(1/3)[0 - 1] =
1/3
4(1/3) = 4/3
You forgot you were evaluating only 1/4 of the area.

Answer 2

After the subsitution for 1-x^2, the limits of u will change from 1 to 0.

I can't find anything wrong with your steps. (I am assuming that the hint you have given is correct)

If you are expecting the entire area bounded by the curve to be 4/3, then you only need to multiply the area bounded by the one-fourth portion with 4.
ie 4 *1/3