Given an integral $$\iint\limits_\triangle f(x,y) ~dx~dy$$ switch the variables to polar coordinates, where $\triangle = \{\text{Triangle whose vertices are at } (1,0),(0,2),(5,3)\}$
Please see the edit at the bottom
(Attached the hand drawn figure below).
I'm taking the substitution $$x = 1 + r\cos\theta, \qquad y = r\sin\theta$$ that is a circle centred at $(1,0)$. As shown in the figure, one can fill one part of triangle with circles (say the part $\Omega_1$).
- In $\Omega_1$, we have $$0 \le r \le \frac{7}{\sqrt{26}}$$ (where $|AH|=\frac{7}{\sqrt{26}})$ and $$\arctan \frac 34 \le \theta \le \pi - \arctan 2$$ that is the range of angle between the sides $AB$ and $AC$. $$\iint\limits_{\Omega_1} f(x,y)~dx~dy = \int\limits_{\arctan \frac 34}^{\pi - \arctan 2} d\theta \int\limits_{0}^{7/\sqrt{26}} f(1+r\cos\theta, r\sin\theta)|J|dr$$ Well, $\Omega_1$ is done.
- Now let $\Omega_2$ be the unfilled region with circles which contains the point $B$ (upper-right region) and $\Omega_3$ be the unfilled region with circles which contains the point $C$ (upper-left region).
- While dealing with these regions, however, $r$ and $\theta$ seems to be "dynamic" in the sense that their bounds are dependent to the other variable.
- I'm not sure how to proceed.
Edit:
With the following new figure, I seem to have a better approach.
Then $$\iint\limits_\triangle f(x,y) ~dx~dy = \int_{\arctan(3/4)}^{\pi-\arctan(2)} d\theta \int_0^{7/(5\sin\theta - \cos\theta)} f(1+r\cos\theta, r\sin\theta)r ~dr$$ since $$BC: y = \frac{1}{5}x+2 \\ \implies r\sin\theta = \frac15(r\cos\theta + 1) + 2 \\ \implies r = \frac{7}{5\sin\theta - \cos\theta}$$ Is it correct?
Your revised approach is correct but you have made some mistakes in your bounds.
If we use substitution in polar coordinates as, $x = 1 + r\cos\theta, y = r\sin\theta$ (in essence we shift the origin to $(1,0)$) -
Measuring the radial distance ($r$) from point $A$, $r$ is bound above by line segment $BC \ (5y-x=10)$ whereas angle $\theta$ is bound between line segments $AB \ (3x-4y = 3)$ and $AC \ (2x + y = 2)$.
$5y -x = 10 \implies 5r\sin\theta - (1 + r\cos\theta) = 10$
That gives upper bound of $r = \dfrac{11}{5\sin\theta - \cos\theta}$
You found the bounds of $\theta$ correctly.
So the integral should be,
$\displaystyle \int_{\arctan(3/4)}^{\pi - \arctan(2)} \int_0^{11/(5\sin\theta - \cos\theta)} f(1+r\cos\theta, r\sin\theta) \ r \ dr \ d\theta$