I'm supposed to express each double integral over the given region R as an iterated integral in polar coordinates

The finite region bounded by the y-axis, the line y = a, and a quarter of the circle of radius a and center at (a, 0)

I tried evaluating this by shifting the origin to $(0, a)$ ranging $\theta$ from $\frac {3\pi}{2}$ to $2\pi$ and $r$ from $0$ to $2a cos\theta$.

$\int_{\frac {3\pi}{2}}^{2\pi} \int_{0}^{2a cos\theta} r dr d\theta$

Is this wrong? how do I shift the origin to (0, a) and evaluate this? or is this not possible?

The solution given is

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{2acos\theta}^{a cosec\theta} r dr d\theta$

considering that r ranges from $2a cos\theta$ to $r sin \theta =a$ i.e $r= a cosec\theta$ I can't seem to understand these bounds since these include part of the circle.

I'm really confused here!

  • 1
    $\begingroup$ If under "shifting" you understand change of order of integration (en.wikipedia.org/wiki/Order_of_integration_(calculus)), then in place of $dr d\theta$ you need to obtain $d\theta dr$, which is not your second integral. Can you explicitly formulate what you mean under "shift coordinates"? $\endgroup$
    – zkutch
    Commented Aug 1, 2023 at 6:39
  • $\begingroup$ what I mean by shift cor-ordinates is shift the origin to (0, a) and evaluate this. $\endgroup$
    – Orpheus
    Commented Aug 1, 2023 at 6:42
  • $\begingroup$ If I correctly understand, then you need linear change of variables in the integral (en.wikipedia.org/wiki/…) - correct? $\endgroup$
    – zkutch
    Commented Aug 1, 2023 at 6:53
  • $\begingroup$ If you are "supposed to express each double integral ... in polar coordinates" your angles cannot just be in the range $3\pi/2$ to $2\pi\,.$ $\endgroup$
    – Kurt G.
    Commented Aug 1, 2023 at 7:00
  • $\begingroup$ why can't my angle be from $\frac{3\pi}{2}$ to $2\pi$ $\endgroup$
    – Orpheus
    Commented Aug 1, 2023 at 7:02

2 Answers 2


It is possible to shift the origin from O $=(0, 0)$ to O' $= (a, 0)$ and setup the integral based on your new origin. However, the results won't be the same until after you fully integrate both the textbook's solution and your solution. Using Wolfram Alpha to check if the integrals are equal is an exercise to possibly practice on.

However, the inner most limits of integration for r are incorrect for the desired area. In polar coordinates, a circle centered at the origin has equation $r = a$. So the lower, initial limit of integration is a. Describing the upper limit of integration becomes a tad problematic because it is the union of two line segments. Thankfully, the region is symmetric along the ray $\theta = \frac{3\pi}{4}$. I'll choose the line along the top, which has equation $r = sec \space \theta$. Then double the area of the top potion to get both the top portion and the left portion.

2$\int_{\frac {3\pi}{4}}^{\pi} \int_{a}^{2a\space sec\theta} r dr d\theta$

Finally, consider a technicality. For the moment ignore that we had to split the area in half and double the integral. This paragraph will only examine entire quadrants. The $\theta$ could then range from $\frac{\pi}{2}$ to $\pi$ because the area is in quadrant II, not $\frac {3\pi}{2}$ which describes quadrant IV. In this exercise because of symmetry of a circle it should not matter, but in other exercises it will matter.

Note: One way to check the integral is setup correctly. Geometry tells us the same area. Remember, the area of the square is $a^2$. The area of 1/4 of the circle is $\frac14 \pi a^2$. Subtracting the two results in $$(1- \frac{\pi}{4}) \cdot a^2$$ which is the answer Wolfram Alpha gives for both the integral when shifted and the textbook's solution.

  • $\begingroup$ Thanks a lot @nickalh after quite some pondering I realised that r value I need can be obtained by simply obliterating the region inside the circle by considering the range for r from $acos \theta$ from the boundary of the circle to the line $y=a$ that gives $r=acosec \theta$ and ranging $\theta$ from $\frac{pi}{4}$ to $\frac{\pi}{2}$ earlier I was partly confused coz I thought that these bounds included some part of the circle…..so I decided to change the origin and evaluate and then I got lost in the mess of the origin shifting to $(0,a)$….anyways I guess you shifted the origin to $(a,0)$ $\endgroup$
    – Orpheus
    Commented Aug 1, 2023 at 10:35
  • $\begingroup$ Anyways thanks a lot @nickalh for the evaluation using shifting of origins $\endgroup$
    – Orpheus
    Commented Aug 1, 2023 at 10:37
  • $\begingroup$ Oops, I just realized you shifted to a different origin than I did. While it's possible to use that origin, why would you use that origin? And by acosec you must mean arc cosecant? or do you mean $a \cdot cosecant \theta = a \cdot csc \theta $ $\endgroup$
    – nickalh
    Commented Aug 1, 2023 at 12:03
  • $\begingroup$ I mean $a cosec at \theta$ as in $a . cosecant \theta$….thanks a lot for evaluating with origin shift….I solved it out and got the same result myself $\endgroup$
    – Orpheus
    Commented Aug 1, 2023 at 13:03

Here is a picture courtesy of Desmos that explains how the integral should be set up with $dr$ first (with $a=1$)

enter image description here

The colored lines show the rays that are integrated along for $dr$ integrals, and for each of the lines, they enter the green region from the circle and exit the green region from the straight line above. This means the integral should be set up as

$$\int\limits_{\text{circle = line}}^{\frac{\pi}{2}}\:\int\limits_{\text{ circle}}^{\text{line}}r\:dr\:d\theta$$

The equation of the circle in polar coordinates is

$$\begin{cases}(x-a)^2+y^2 = a^2 \\ x^2+y^2=2ax\end{cases} \implies \begin{cases}r^2 = 2ar\cos\theta \\ r = 2a\cos\theta\end{cases}$$

and the line is

$$y = a \implies \begin{cases}r\sin\theta = a \\ r = a\csc\theta\end{cases}$$

Finally, the circle and the line intersect at

$$2a\cos\theta = a\csc\theta \implies \sin2\theta = 1$$

or $\theta = \frac{\pi}{4}$. This gives the integral from your answer key.


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