# Iterated Integral

1. Feb 9, 2009

### cse63146

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

Evaluate the following double integral over the closed region B:

$$\int\int_B x^2 y^3 dx dy$$ where B is the region bounded by y = x^2 and y = x

2. Relevant equations

3. The attempt at a solution

I think I set up the paramaters wrong:

$$\int^{1}_{0}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy$$

Last edited: Feb 9, 2009
2. Feb 9, 2009

### Dick

If y is in [0,1], as it looks like you've figured out for your integral, which is larger y or sqrt(y)? Is that what's confusing you?

3. Feb 9, 2009

### cse63146

y>sqrt(y)

The answer is 1/77, but I get -1/77, but taking it's absolute value should take care of that.

4. Feb 10, 2009

### Unco

What's sqrt(1/4)?

5. Feb 10, 2009

### HallsofIvy

Staff Emeritus
Can you explain why "taking its absolute value" is justified?

Once again, which is larger, y or $\sqrt{y}$ for y between 0 and 1?

Which is larger 1/4 or $\sqrt{1/4}$?

6. Feb 10, 2009

### cse63146

I get it. Because y goes from 0 - 1, sqrt(y)>y in this case.

but what if y went from 0 - 2, then would it be this:

$$\int^{1}_{0}\int^{\sqrt{y}} }_{y} x^2 y^3 dx dy + \int^{2}_{1}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy$$

7. Feb 10, 2009

### Dick

Yes.

8. Feb 10, 2009

### cse63146

Thank you all for your help.