Consider the integrals:

$$ I=\int_{0}^{\sqrt{2}}\int_{y^2}^{2}y\ dxdy \\ I'=\int_{y^2}^{2}\int_{0}^{\sqrt{2}}y\ dydx $$

From what I understand, the order of integration does not matter. However, as can be easily shown by hand or a simple online integration software:

$$I=1 \\ I'=2-y^2$$

Why are these not the same?

  • 2
    $\begingroup$ Draw a picture of the region you are integrating over. Then you have a sporting chance of understanding what you did wrong. A rule: only the limits of the inner integral may depend on the variable. The limits of the outer integral must be constants. $\endgroup$ Commented Jan 20, 2019 at 10:19

1 Answer 1


You can reverse the order but the limits of integration change: $$\int_{y=0}^{\sqrt{2}}\left(\int_{x=y^2}^{2}y\ dx\right)dy=\int_{x=0}^{2}\left(\int_{y=0}^{\sqrt{x}}y\ dy\right)dx.$$ enter image description here

  • 3
    $\begingroup$ I took the liberty of adding a picture of the region the integral is over. My experience is that students confused with iterated integrals too often fail to do that. Mind you, their confusion does not end there, but my thesis is that without the picture (at least in their minds) they have no chance. Anyway, if you object, just ping me, and I will remove it. $\endgroup$ Commented Jan 20, 2019 at 10:15
  • $\begingroup$ @JyrkiLahtonen A picture here helps a lot. Thanks! $\endgroup$
    – Robert Z
    Commented Jan 20, 2019 at 10:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .