Find the area of the region bounded by the curves y=x^2 and x=y^2

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Find the area of the region bounded by the curves y=x^2 and x=y^2

Mathematics
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double integral it ;)
equal both equation to each other to find the boundaries:\[x^2 = x^{1\over2} \] Area of a bounded region:\[\int\limits_{a}^{b}Ytop - Ybottm dy\] where Ytop = which curve is the top part of the region Ybottom = curve that is the bottom part of the region a & b are the boundaries you found
the curves intersect at (0,0) and (1/1) .... integrate x^(1/2) - x^2 between x=0 and x=1

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or simply see that the area is the region bound by the y = sqrt(x) and y=x^2 curves
aash got it ;)
Math did too lol
alright thanks, know how to do regions between curves, just the x=y^2 is new and strange to me
thanks amistre63
whenever you have something like x=y^2 just solve for y to have a "normal" equation x=y^2 ==> y=x^(1/2)
x=y^2 is same as y=x^(1/2)
if we double integral it we would have to define a region for a domain and use z=0 as a starting point ;)
y=+ or - sqrt(x) though doesnt it
ther are 2 points of intersection of functions- (0,0) and (1,1) Therefore area is\[\int\limits_{0}^{1}(\sqrt{x}-x^2)=1/3\]
Ty; yes; but the -sqrt(x) is useless
for area take + only
area cannot be negative
the left side of x^2 and the bottom of y=x^2 are pointless ;)
if you get a negative area; take the absolute value to asdjust for errors in the way to subtracted
ah ok thanks a lot everyone,
haha and 1+1=2 :D
the real reason for not taking the negative root is to make it a function
5-3 = 2 3-5 = -2 ...just means you put them in the wrong order usually; so: |+-2| = 2
ok great, thanks again

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