Double integrals interchanging order

[jonroberts74]

Homework Statement

$$\int_{1}^{4}\int_{1}^{\sqrt{x}}(x^2+y^2)dydx$$

The Attempt at a Solution

I drew the region,

I tried
$$\int_{1}^{2}\int_{1}^{y^2}(x^2+y^2)dxdy$$

but it doesn't seem to work.

when the order is changed

$$1 \le y \le 2$$

and $$\sqrt{x} = y \rightarrow x=y^2$$

so the change in x is $$1 \le x \le y^2$$

but it doesn't work out to the correct answer.

 

[slider142]

The region described by $1\leq y \leq 2$ and $1\leq x\leq y^2$ looks like this:

However, the region specified in the first integral is this one:

Can you tell what the proper bounds should be from this illustration ?

 

[jonroberts74]

I see it

the new integral should be

$$\int_{1}^{2}\int_{y^2}^{4} (x^2+y^2)dxdy = \int_{1}^{4}\int_{1}^{\sqrt{x}} (x^2+y^2)dydx$$

 

[slider142]

That's it. :)