# Change in order of integration (Polar coordinates)

How does one change the order of integration in polar coordinates?

For example: Change the order of integration of following integral $$\int_0^{\frac{\pi}{2}}\int_0^{2a \cos \theta} f(r,\theta) \,dr\, d\theta$$

I know how to change the order of integration when we are working in rectangular coordinates, but I am unable to visualize this.

I was looking for some books/articles or some explanatory things to visualize this. Thanks in advance.

Edit

Changing the order of double integrals in polar coordinates.

I found one similar question here but I didn't get the idea. It would be good if somebody can explain that to me.

• There is an $r$ missing in your integral, which I think, is also the issue in your previous post, where I commented. See example 2 in the following link: tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx Oct 1, 2017 at 18:30
• $r$ is not missing. I am not converting rectangular coordinates to polar coordinates.
– User
Oct 2, 2017 at 0:50
• @imranfat, I don't find any relation between example 2 there and my question.
– User
Oct 2, 2017 at 0:55
• Did you check example 2? That is exactly a polar coordinates graph problem with area between two curves. Read after "The area of the region D is then...". There is an $r$ and that is the only way how you can get these fractional answers in your earlier post, instead of $\pi$.It is also clear in your link after EDIT Oct 2, 2017 at 3:27
Firstly, it helps to analyze the region suggested by $$0\le\theta\le\pi/2$$ and $$0\le r\le2a\cos\theta$$. Let's focus on the boundary $$r=2a\cos\theta$$ and do some algebra: $$r=2a\cos\theta=2a\left(x/r\right)$$, so $$r^{2}=2ax$$. Thus, $$x^{2}-2ax+y^{2}=0$$. Completing the square, we have $$\left(x-a\right)^{2}-a^{2}+y^{2}=0$$, so that $$\left(x-a\right)^{2}+y^{2}=a^{2}$$. This is the circle of radius $$\sqrt{a^{2}}=\left|a\right|$$ centered at $$\left(a,0\right)$$.
Now we need to understand which portion of this circle we get with $$0\le\theta\le\pi/2$$. To simplify things, I'm going to assume $$a>0$$. At $$\theta=0$$, $$r$$ goes from $$0$$ to $$2a$$, so we get the entire diameter on the $$x$$-axis. At $$\theta=\pi/4$$, $$\cos\theta=\sqrt{2}/2$$, and $$r=2a\cos\theta=a\sqrt{2}$$; this is the point $$\left(x,y\right)=\left(a,a\right)$$. And as $$\theta$$ increases towards $$\pi/2$$, $$r=2a\cos\theta$$ continues to decrease towards $$0$$. $$r$$ never becomes negative, so this is entirely in the first quadrant, and must be the upper semi-circle, like $$\smallfrown$$.
Now, if we wanted to change the order of this integration, we would need something like $$\int_{r_{1}}^{r_{2}}\int_{\theta_{1}\left(r\right)}^{\theta_{2}\left(r\right)}f\left(r,\theta\right)\,\mathrm{d}\theta\,\mathrm{d}r$$. So what are the absolute bounds for $$r$$? $$2a\cos\theta$$ is maximized for $$0\le\theta\le\pi/2$$ when $$\theta=0$$, so the most $$r$$ can be is $$2a$$. The least it can be is $$0$$, just as in the original iterated integral. So we have $$\int_{0}^{2a}\int_{\theta_{1}\left(r\right)}^{\theta_{2}\left(r\right)}f\left(r,\theta\right)\,\mathrm{d}\theta\,\mathrm{d}r$$.
Now we need to know the bounds for $$\theta$$, given a particular value of $$r$$. This is a little weird to picture since a value of $$r$$ fixes a circle centered at the origin and $$r=2a\cos\theta$$ is a different circle not centered at the origin. But we don't need to visualize it. All values of $$r$$ from $$0$$ to $$2a$$ are included for $$\theta=0$$ by the discussion earlier, so $$0$$ is always the minimum value of $$\theta$$. The maximum value of $$\theta$$ would be on the boundary semicircle $$r=2a\cos\theta$$. To get $$\theta$$ from $$r$$, we just solve: $$\theta=\arccos\left(\dfrac{r}{2a}\right)$$. When $$0\le r\le2a$$, this gives a value in $$\left[0,\pi/2\right]$$, as desired. Therefore, our bounds for $$\theta$$ are $$0$$ and $$\arccos\left(\dfrac{r}{2a}\right)$$. And the final iterated integral is $$\int_{0}^{2a}\int_{0}^{\arccos\left(r/(2a)\right)}f\left(r,\theta\right)\,\mathrm{d}\theta\,\mathrm{d}r\text{.}$$