SRL Archive - Mathematics

Hiya,
I hope some of you here are good at calculus and are a bit further ahead than me, cause I need some example of determining the bounds of a double integral.
Adams 6th/7th edition 14.2 excercise 9:
, where R is the finite region in the first quadrant bounded by the curves y = x² and x = y²

The solutions manual goes straight to this:
My question: how did they determine the x² and the sqrt(x)?

The first curve is y=x^2, so it is the first boundary of R. The second boundary is, as you said, y^2=x, which can be also stated as y=+/-sqrt(x). You use y=+sqrt(x) because it's the first quadrant.

Dang, never looked at it that way. Thanks a lot!
For future reference: 0 to 1 is ofc because y=x² (and y² = x) has solutions x = 0 and x = 1

Edit: follow-up question: why from x² to sqrt(x) and not other way around?
Edit2: obviously because for x < 1, sqrt(x) is larger than x², god I'm in bad shape.

From: Markus
Dang, never looked at it that way. Thanks a lot!
For future reference: 0 to 1 is ofc because y=x² (and y² = x) has solutions x = 0 and x = 1

Edit: follow-up question: why from x² to sqrt(x) and not other way around?
Edit2: obviously because for x < 1, sqrt(x) is larger than x², god I'm in bad shape.

because on range 0-1 sqrt(x) >= (x) >= x^2

z

edit: and as it has been already said y=x^2 is one bounding element and y^2=x which is equivalent to y=sqrt(x) is the other bounding element