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

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