Is $\int_0^2 [\int_x^ {\sqrt 3x} f(\sqrt{x^2+y^2})dy]dx=\int_{\pi /4}^{\pi /3}[\int_0^{2\sec \theta}rf(r)dr]d\theta$?

Question

Is it true that $\int_0^2 [\int_x^ {\sqrt 3x} f(\sqrt{x^2+y^2})dy]dx=\int_{\pi /4}^{\pi /3}[\int_0^{2\sec \theta}rf(r)dr]d\theta$ ? . I can't figure out what polar co-ordinate transformation would it be and how to determine the limits of integration in the changed system . Please help . Thanks in advance

Answer 1

Sketch the region of integration: $$ R = \{(x, y) \in \mathbb R^2 \mid x \in [0, 2] \text{ and } y \in [x, \sqrt 3 x]\} $$ Observe that it is a triangle with vertices $(0, 0)$, $(2, 2)$ and $(2, 2\sqrt 3)$. First, we seek the lower and upper angles $\theta_1$ and $\theta_2$. For the lower bound, we see that line $y = x$ can be manipulated into: $$ \tan \theta_1 = \frac{y}{x} = 1 \implies \theta_1 = \frac{\pi}{4} $$ Likewise for the upper bound, the line $y = \sqrt 3 x$ can be manipulated into: $$ \tan \theta_2 = \frac{y}{x} = \sqrt 3 \implies \theta_2 = \frac{\pi}{3} $$ Finally, we analyze the lower and upper bounds for the radius $r$ as a function of $\theta$. From our diagram, we see that the lower bound is simply $r = 0$ and that the upper bound is the vertical line $x = 2$. But since $x = r \cos\theta$, it follows that: $$ r \cos\theta = 2 \iff r = 2\sec\theta $$