# Double integrals interchanging order

1. Jul 22, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

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

3. 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.

2. Jul 23, 2014

### 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 ?

#### Attached Files:

File size:
3.6 KB
Views:
81
• ###### Plot2.png
File size:
3.6 KB
Views:
87
3. Jul 23, 2014

### jonroberts74

I see it

the new integral should be

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

4. Jul 23, 2014

### slider142

That's it. :)