# double integral, how should i determine the integral boundaries ?

#### akleron

##### New member
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

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

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 $$\displaystyle y$$ in the positive direction, as a function of $$\displaystyle y$$
what value of $$\displaystyle x$$ does your area start at? What value does it end at?

Those will be your lower and upper integration limits for $$\displaystyle x$$

#### akleron

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

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 $$\displaystyle y$$ in the positive direction, as a function of $$\displaystyle y$$
what value of $$\displaystyle x$$ does your area start at? What value does it end at?

Those will be your lower and upper integration limits for $$\displaystyle x$$
so the integration limits for x should be from y=x^2 to
y=√x ?

#### akleron

##### New member
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

##### Full Member
so the integration limits for x should be from y=x^2 to
y=√x ?
No.

The area starts at $$\displaystyle y = \sqrt{x} \Rightarrow x = y^2$$

It ends at $$\displaystyle y = x^2 \Rightarrow x = \sqrt{y}$$

Thus the limits are $$\displaystyle x: y^2 \to \sqrt{y}$$

so the overall area is

$$\displaystyle A = \displaystyle \int_0^1 \int_{y^2}^{\sqrt{y}}~dx~dy$$

yeah, you got it right in the next post. Cheers.

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