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

1. amistre64

double integral it ;)

2. anonymous

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

3. anonymous

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

4. amistre64

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

5. amistre64

aash got it ;)

6. amistre64

Math did too lol

7. anonymous

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

8. anonymous

thanks amistre63

9. anonymous

whenever you have something like x=y^2 just solve for y to have a "normal" equation x=y^2 ==> y=x^(1/2)

10. anonymous

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

11. amistre64

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

12. anonymous

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

13. mathmagician

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

14. amistre64

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

15. anonymous

for area take + only

16. anonymous

area cannot be negative

17. amistre64

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

18. amistre64

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

19. anonymous

ah ok thanks a lot everyone,

20. anonymous

haha and 1+1=2 :D

21. anonymous

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

22. amistre64

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

23. anonymous

ok great, thanks again