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This is a general question which I have been asking (to no avail) for a while around my class. I will ask my question using an example.

Evaluate $I=\displaystyle\int_{0}^{1}\mathrm{d}x\displaystyle\int_{\sqrt{x}}^{1}e^{y^3}\mathrm{d}y$.

Where the solution provided by the lecturer is

We cannot directly integrate $e^{y^3}$ with respects to $y$, therefore we express $I$ as a double integral over a region $D$, that is $$I=\iint_{D}e^{y^3}\mathrm{d}x\,\mathrm{d}y.$$ We can describe $D=\{(x,y):0\leqslant y\leqslant 1, 0\leqslant x\leqslant y^2\}$:

which is shown in this picture.

The lecturer goes on to solve the integral using these limits and everything is fine.

My question is: why do we choose $0\leqslant x\leqslant y^2$? What is the difference between this and $y^2\leqslant x\leqslant 1$? The same can be asked for the limits of the first integral: what is the difference between $\sqrt{x}\leqslant y\leqslant 1$ and $0\leqslant y\leqslant \sqrt{x}$?

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  • $\begingroup$ Yes, and I have attached the illustrated region. For me, unfortunately, it does not illuminate much. Both versions of the limits I have presented, in my mind, 'end up' at the same place. That is, for $\sqrt{x}\leqslant y\leqslant 1$, $y$ 'starts at' 0 and 'ends up at' 1, as it does in $0\leqslant y\leqslant\sqrt{x}$. I feel as though my line of thought is wrong, which is why I am struggling to understand why we choose limits in this way. $\endgroup$ – Toby Reichelt May 23 at 12:42
  • $\begingroup$ bounds of integration need to cover the zone of integration $\endgroup$ – Makina May 23 at 12:43
  • $\begingroup$ @TobyReichelt if, for a given $x$, you integrate, wrt $y$, between $0$ and $\sqrt x$, you do not cover the region but the area below it. In order to see this, just pick a value of x in the range from $0$ to $1$ and draw the vertical line with that abscissa. See now which part of the line is inside $D$. $\endgroup$ – Matteo May 23 at 12:45
  • $\begingroup$ Thank you for your response. I think I am starting to understand---since $y^2\leqslant x\leqslant 1$, this would suggest that my region on the $y$ axis is going to be all the $y^2$ 'below' the $x$-values, and for $0\leqslant x\leqslant y^2$, my region on the $y$ axis is going to be all the $y^2$ 'above' the $x$-values. $\endgroup$ – Toby Reichelt May 23 at 12:54
  • $\begingroup$ @TobyReichelt, more precisely: If you invert the order of integration, now, for a given $y$ between $0$ and $1$ you draw the horizontal line. For that ordinate, what is the interval of $x$ over which you must integrate? The answer is between $0$ and the square of the chosen $y$! $\endgroup$ – Matteo May 23 at 13:00
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My question is: why do we choose $0\leqslant x\leqslant y^2$? What is the difference between this and $y^2\leqslant x\leqslant 1$? The same can be asked for the limits of the first integral: what is the difference between $\sqrt{x}\leqslant y\leqslant 1$ and $0\leqslant y\leqslant \sqrt{x}$?

To clarify this, take a look at the region with these extra markings:

enter image description here

For $0 \le y \le 1$, you need to set up the limits for $x$ so that you run through the blue region $D$. Place yourself at a virtual $y$ between $0$ and $1$ and coming from the left (for $x$), notice that the region starts at $x=0$ (the $y$-axis) and ends at the parabola $x=y^2$, indicated by the green line. Each point in the blue region satisfies $0 \le x \le y^2$.

The bounds you suggest as being equally fine, $y^2\leqslant x\leqslant 1$, would refer to points after the parabola but before the line $x=1$, indicated by the red line. Any $x$ satisfying these inequalities is not located in the blue region $D$, but in the (lightly) red shaded region.


The clue is that for a double integral, there are two possible orders of integration: first with respect to $x$ and then $y$; and vice versa. Depending on the region of integration and on the function you wish to integrate, differences between the two orders can arise:

  • integration can be significantly harder/easier in one of the orders, e.g. you may not be able to (easily) find an anti-derivative in one of the integration orders;
  • the region is more easily described in one of the two orders, e.g. because you can do it in one iterated integral without the need of splitting the region into more than one part.

For the region of integration in your example, the order of integration does not really affect how hard it is to set up the boundaries: it can be done in one iterated integral in both orders:

enter image description here

Projection of the blue region $D$ onto the $x$-axis gives $0 \le x \le 1$ and for an arbitrary $x$ in that interval, you have $\sqrt{x} \le y \le 1$. These are the limits in the integral that was initially given. However, this requires finding an anti-derivative of $e^{y^3}$ with respect to $y$...

Projection of the blue region $D$ onto the $y$-axis gives $0 \le y \le 1$ and for an arbitrary $y$ in that interval, you have $0 \le x \le y^2$. The advantage of this order is that you now have to integrate with respect to $x$ first which, with the given function $e^{y^3}$, is easy.

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  • $\begingroup$ I think the OP is having difficulties understanding why the limits of integration are expressed in that way. $\endgroup$ – Matteo May 23 at 12:51
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    $\begingroup$ Yes, I see - I added a part to the top of my answer to address that question. $\endgroup$ – StackTD May 23 at 13:02
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With thanks to Matteo, I have finally understood the issue. Whilst it is unlikely that anyone is facing the same issue, I will post an explanation regardless for the few that might need it.

When we express the limits as $y^2\leqslant x\leqslant 1$, we are covering all the $y^2$ that falls 'below' the corresponding $x$-values.

When we express the limits as $0\leqslant x\leqslant y^2$, we are covering all the $y^2$ that falls 'above' the corresponding $x$-values.

Therefore, for this problem, since we want all the region that's 'above' the corresponding $x$-values, so we choose $0\leqslant x\leqslant y^2$.

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