double integral, how should i determine the integral boundaries ?

[akleron]

I need to calculate the following integral shown in this picture.
I think I managed to draw the area of D correctly [shown in the picture ] but i don't know how to continue..any help ?

 

[Romsek]

$\displaystyle x^2 = \sqrt{x} \Rightarrow x = 1$

So it should be pretty clear that the outer integration limits, for $\displaystyle y$ are $\displaystyle 0 \to 1$

Now look at your graph. As you move [MATH]y[/MATH] in the positive direction, as a function of [MATH]y[/MATH]what value of [MATH]x[/MATH] does your area start at? What value does it end at?

Those will be your lower and upper integration limits for [MATH]x[/MATH]

 

[akleron]

$\displaystyle x^2 = \sqrt{x} \Rightarrow x = 1$

So it should be pretty clear that the outer integration limits, for $\displaystyle y$ are $\displaystyle 0 \to 1$

Now look at your graph. As you move [MATH]y[/MATH] in the positive direction, as a function of [MATH]y[/MATH]what value of [MATH]x[/MATH] does your area start at? What value does it end at?

Those will be your lower and upper integration limits for [MATH]x[/MATH]

so the integration limits for x should be from y=x^2 to
y=√x ?

 

[akleron]

I think I got the idea now.
It's still hard for me to determine the integration boundaries but I guess that I will figure it out ...
Thanks for the help !

 

[Romsek]

so the integration limits for x should be from y=x^2 to
y=√x ?

No.

The area starts at [MATH]y = \sqrt{x} \Rightarrow x = y^2[/MATH]
It ends at [MATH]y = x^2 \Rightarrow x = \sqrt{y}[/MATH]
Thus the limits are [MATH]x: y^2 \to \sqrt{y}[/MATH]
so the overall area is

[MATH]A = \displaystyle \int_0^1 \int_{y^2}^{\sqrt{y}}~dx~dy[/MATH]
yeah, you got it right in the next post. Cheers.