Polar Coordinates Double Integration

Question

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!

Answer 1

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.

Answer 2

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

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.