Question

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

Answer 1

double integral it ;)

Answer 2

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

Answer 3

the curves intersect at (0,0) and (1/1) .... integrate x^(1/2) - x^2 between x=0 and x=1

Answer 4

or simply see that the area is the region bound by the y = sqrt(x) and y=x^2 curves

Answer 5

aash got it ;)

Answer 6

Math did too lol

Answer 7

alright thanks, know how to do regions between curves, just the x=y^2 is new and strange to me

Answer 8

thanks amistre63

Answer 9

whenever you have something like x=y^2 just solve for y to have a "normal" equation

x=y^2 ==> y=x^(1/2)

Answer 10

x=y^2 is same as y=x^(1/2)

Answer 11

if we double integral it we would have to define a region for a domain and use z=0 as a starting point ;)

Answer 12

y=+ or - sqrt(x) though doesnt it

Answer 13

ther are 2 points of intersection of functions- (0,0) and (1,1) Therefore area is\[\int\limits_{0}^{1}(\sqrt{x}-x^2)=1/3\]

Answer 14

Ty; yes; but the -sqrt(x) is useless

Answer 15

for area take + only

Answer 16

area cannot be negative

Answer 17

the left side of x^2 and the bottom of y=x^2 are pointless ;)

Answer 18

if you get a negative area; take the absolute value to asdjust for errors in the way to subtracted

Answer 19

ah ok thanks a lot everyone,

Answer 20

haha and 1+1=2 :D

Answer 21

the real reason for not taking the negative root is to make it a function

Answer 22

5-3 = 2

3-5 = -2 ...just means you put them in the wrong order usually; so:

|+-2| = 2

Answer 23

ok great, thanks again