How can I find the limits of this iterated polar integration?

Question

How can compute the area of the triangle whose corners are at the origin, (1,0) and (1,1). I solved this with r integral first but I could not find the correct limits for theta integral first order. One attempt/observation is that. There must be 2 seperated integrals. It must be something like; [\int \int rd\Theta dr] + [\int \int rd\Theta dr]

Answer 1

Hint: The vertical line segment between $(1,0)$ and $(1,1)$ has equation $r=\sec \theta$ for $0\leq\theta\leq\pi/4$.

So the area can be written as the integral $$A=\int_{\theta=0}^{\pi/4}\int_{r=0}^{\sec\theta}r\; dr\; d\theta$$

You should be able to integrate this directly as an iterated integral.

Note: You probably know that if you draw a ray from the origin at angle $\theta$ to the $x$-axis, then the intersection of the ray with the unit circle gives you the sine and cosine of that angle. What is less commonly memorized is that the intersection of that ray with the vertical tangent to the unit circle gives the tangent of the angle and the secant of the angle. The tangent of the angle is the height of the intersection point and the secant is the distance from the intersection point to the origin.

Answer 2

Given the region of integration $$R = \{(x,y) \in \mathbb R^2 : (0 \le x) \cap (0 \le y) \cap (0 \le x + y \le 1) \},$$ suppose we wish to express the area of $R$ in the form $$\int_{r=?}^? \int_{\theta=?}^? r \, d\theta \, dr$$ where $$(x,y) = (r \cos \theta, r \sin \theta),$$ or equivalently, $$r = \sqrt{x^2+y^2}, \quad \theta = \tan^{-1} \frac{y}{x},$$ where $r \ge 0$ and $-\pi < \theta \le \pi$. Because $R$ lies in the first quadrant, we note that the lower limit of integration for $r$ is obviously $0$. The upper limit corresponds to the maximum distance of a point in $R$ from the origin, namely $r = 1$.

For a given radius $r$--that is to say, for all points in $R$ whose distance from the origin is $r$, we must determine the set of angles corresponding to that radius. Clearly, if $r \in [0, 1/\sqrt{2}]$, $\theta \in [0,\pi/2]$: this is because a quarter circle of radius $1/\sqrt{2}$ can be inscribed in the triangle $R$. But if $1/\sqrt{2} < r \le 1$, then the set of angles $\theta$ is disjoint, comprising two intervals of equal length and symmetric about the line $\theta = \pi/4$. The symmetry of $R$ lets us compromise and write $$|R| = 2 \int_{r = 0}^1 \int_{\theta = 0}^{g(r)} r \, d\theta \, dr,$$ where $$g(r) = \frac{\pi}{4} - \begin{cases} \cos^{-1} \frac{1}{r \sqrt{2}}, & \frac{1}{\sqrt{2}} < r \le 1 \\ 0, & 0 \le r \le \frac{1}{\sqrt{2}}. \end{cases}$$