# Double integrals help

1. Jul 29, 2014

### jonroberts74

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

Set up the double integral over the region $y=x+3; y=x^2+1$

2. Relevant equations

3. The attempt at a solution

finding the intersections you get the double integral

$\int_{1}^{5}\int_{-1}^{2}dxdy =12$

but why is that not the same as

$\int_{1}^{5}\int_{-1}^{\sqrt{y-1}} dxdy =\frac{28}{3}$

2. Jul 29, 2014

### LCKurtz

That would be the integral over a rectangle.

That's wrong too. Have you drawn the graphs?

3. Jul 29, 2014

### jonroberts74

yeah I drew it, its the line y=x+3 slicing through the parabola x^2 moved up one unit on the y-axis

$\int_{1}^{5}\int_{-\sqrt{y-1}}^{\sqrt{y-1}} dxdy$ is the other one I had, but no sure if that is correct either

4. Jul 29, 2014

### LCKurtz

It isn't. If you do x first it must go from $x_{left}$ to $x_{right}$. x on the left is a two piece formula.

Is there some reason you don't do this problem the more natural and easier dydx order?

5. Jul 29, 2014

### jonroberts74

probably out of habit of typically slicing with a y value then seeing how x changes

$\displaystyle\int_{-1}^{2}\int_{x^2+1}^{x+3} dydx$

okay I see it now, if I do x first theres an issue where the left intersection point occurs, below that point the parabola is the left and right x and above that the left is the line and the right is the parabola.

thanks!

6. Jul 29, 2014

### LCKurtz

Yes! That's why you need two integrals to do it it dxdy order. That's the sort of thing you should always look at when deciding how to set up a double integral. Now can you get the same answer both ways?

7. Jul 29, 2014

### jonroberts74

$\displaystyle \int_{1}^{2} \int_{-\sqrt{y-1}}^{\sqrt{y-1}}dxdy + \int_{2}^{5} \int_{y-3}^{\sqrt{y-1}}dxdy = \frac{4}{3} + \frac{19}{6} = \frac{9}{2}$