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anonymous
 5 years ago
Find the area of the region bounded by the curves y=x^2 and x=y^2
anonymous
 5 years ago
Find the area of the region bounded by the curves y=x^2 and x=y^2

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1double integral it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0equal 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the curves intersect at (0,0) and (1/1) .... integrate x^(1/2)  x^2 between x=0 and x=1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1or simply see that the area is the region bound by the y = sqrt(x) and y=x^2 curves

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright thanks, know how to do regions between curves, just the x=y^2 is new and strange to me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whenever you have something like x=y^2 just solve for y to have a "normal" equation x=y^2 ==> y=x^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=y^2 is same as y=x^(1/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if we double integral it we would have to define a region for a domain and use z=0 as a starting point ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=+ or  sqrt(x) though doesnt it

mathmagician
 5 years ago
Best ResponseYou've already chosen the best response.0ther are 2 points of intersection of functions (0,0) and (1,1) Therefore area is\[\int\limits_{0}^{1}(\sqrt{x}x^2)=1/3\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1Ty; yes; but the sqrt(x) is useless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0area cannot be negative

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the left side of x^2 and the bottom of y=x^2 are pointless ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if you get a negative area; take the absolute value to asdjust for errors in the way to subtracted

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah ok thanks a lot everyone,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the real reason for not taking the negative root is to make it a function

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.153 = 2 35 = 2 ...just means you put them in the wrong order usually; so: +2 = 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok great, thanks again
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