# Tried reverse engineering this Double Integral problem, but it left me more confused

I feel like this problem should be really easy, but I feel like every attempt I make to solve it goes nowhere.

The problem is to convert the following integral to polar coordinates: $$\int_0^{\frac12}\int_0^{1-x}dydx = \frac38$$

Which has the easy-to-compute value of $$\frac38$$. This makes it easy to verify any attempt at an equivalent polar integral, because all I have to do is make sure that its area is the same as the original.

To begin, I was taught that a good way to start with problems like this is to graph the region that is being integrated:

With this, it's not hard to see that $$0 < \theta < \frac{\pi}{2}$$. However, the range for $$r$$ has been a little more challenging. All we know is $$r > 0$$, and that it is bounded by some unknown upper function of $$\theta$$, $$f(\theta)$$.

We know that $$0 < x < \frac12$$ and $$0 < y < 1 - x$$, so substituting the appropriate trig identities gives $$0 < r\cos(\theta) < \frac12$$ and $$0 < r\sin(\theta) < 1 - r\cos(\theta)$$.

I've tried many ways to manipulate these inequalities to yield something useful, but the best answer I could get is $$0 < r < \frac{1}{\cos{\theta} + \sin(\theta)}$$. According to my testing on Desmos, it seems that this upper bound is partially correct, in that it does contain the desired area, but it contains more area on top of that.

In a last ditch effort, I decided to "cheat," and reverse engineer the problem by setting up the following equation to solve for the upper bound: $$\frac38 = \int_0^{\frac{\pi}{2}}\int_0^? r\mathrm{d}r\mathrm{d}\theta$$

Where the ? is the unknown upper bound that I'm solving for. Solving this integral equation yielded an upper bound of $$? = \sqrt{\frac{3}{2\pi}}$$.

The problem is, I would have never found this same value had I continued with my original approach. With each attempt, it seemed like I was getting some function of $$\theta$$ as an upper bound--I would have never expected a constant!

Part of me wants to just leave this problem as is, since I have technically already solved it, but it bothers me that I haven't been able to use my "cheat" answer to get the upper bound using the other, more traditional method.

In a nutshell, what I am asking for is help finding this answer through my original method.

P.S., through testing, I stumbled upon the fact that $$? = \frac{\sqrt{3}}{2} \frac{1}{\cos{\theta} + \sin{\theta}}$$ is also a valid solution to this integral. Similar to my other "cheat" answer, though, I'm at a loss at what to do with this answer.

• The upper bound for $r$ is going to be different for $0 < \theta < \pi/4$ and for $\pi/4 < \theta < \pi/2$, so you should split the integral for $\theta$ into those two parts. Apr 27, 2023 at 18:09

In order to think in polar coordinates, picture a radar dish at the origin, sweeping out the region. Clearly, the angle coordinate must cover the entire first quadrant, so $$0 \leq \theta \leq \frac{\pi}{2}$$. However, the radial distance to the outer boundary is best described by a piecewise function of the angle, behaving one way on the first half, then a different way on the second.

For $$0 \leq \theta \leq \frac{\pi}{4}$$, the outer boundary is $$x = \tfrac12$$, or $$r\cos\theta = \tfrac12$$, or $$r = f_1(\theta) = \frac{1}{2\cos\theta} = \frac12\, \sec\theta.$$

For $$\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$$, the outer boundary is $$y = 1 - x$$, or $$r\sin\theta = 1 - r\cos\theta$$, or $$r = f_2(\theta) = \frac{1}{\sin\theta + \cos\theta}.$$ In other words, we have the piecewise function $$f(\theta) = \begin{cases} f_1(\theta), & 0 \leq \theta \leq \frac{\pi}{4} \\ f_2(\theta), & \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2} \end{cases}$$

This piecewise function $$r = f(\theta)$$, plotted in rectangular coordinates, looks like this:

Now, the area integral in polar coordinates is
$$A = \int_0^{\pi/2} \int_0^{f(\theta)} r \, \mathrm{d}r \, \mathrm{d}\theta = \frac12 \int_0^{\pi/2} \bigl[ f(\theta) \bigr]^2 \, \mathrm{d}\theta,$$ which must be split into two pieces since $$f(\theta)$$ has a piecewise definition over the domain: \begin{align} A &= \frac12 \int_0^{\pi/4} \bigl[ f_1(\theta) \bigr]^2 \, \mathrm{d}\theta + \frac12 \int_{\pi/4}^{\pi/2} \bigl[ f_2(\theta) \bigr]^2 \, \mathrm{d}\theta \\ &= \frac18 \int_0^{\pi/4} \sec^2 \theta \, \mathrm{d}\theta + \frac12 \int_{\pi/4}^{\pi/2} \frac{1}{(\sin\theta + \cos\theta)^2} \, \mathrm{d}\theta, \end{align} which can be integrated to give a total area of $$\frac38$$, as expected.

Since:

$$D = \left\{(x,y)\in\mathbb{R}^2 : 0 \le x \le \frac{1}{2},\,0 \le y \le x\right\} \cup \left\{(x,y)\in\mathbb{R}^2 : 0 \le x \le \frac{1}{2},\,x \le y \le 1-x\right\}$$

then:

$$D^* = \left\{(\rho,\theta)\in[0,\infty) \times [0,2\pi) : 0 \le \theta \le \frac{\pi}{4},\,0 \le \rho \le \frac{1}{2\cos\theta}\right\} \cup \\ \left\{(\rho,\theta)\in[0,\infty) \times [0,2\pi) : \frac{\pi}{4} \le \theta \le \frac{\pi}{2},\,0 \le \rho \le \frac{1}{\cos\theta+\sin\theta}\right\}.$$

Of course, we can also write:

$$D^* = \left\{(\rho,\theta)\in[0,\infty) \times [0,2\pi) : 0 \le \theta \le \frac{\pi}{2},\,0 \le \rho \le \min\left(\frac{1}{2\cos\theta},\,\frac{1}{\cos\theta+\sin\theta}\right)\right\}$$

but, from my point of view, it's like sweeping the dust under the rug!