Answer:

Explanation:
Evaluate the given double integral bounded by the region made up of the circle, x²+y²=36, and the lines x=0 and y=x, using polar coordinates.

(1) - Changing to polar coordinates from rectangular

Limits:
r^2=36 => r=±6
∴ 0≤r≤6
and π/4≤θ≤π/2
Integrand:

We now have,

(2) - Evaluating the integral
![\int\limits^(\pi/2)_(\pi/4) \int\limits^6_0{( 2r^2\cos(\theta)-r^2\sin(\theta))} \, drd\theta\\\\\\\Longrightarrow \int\limits^(\pi/2)_(\pi/4)\Big[(2)/(3)r^3\cos(\theta)-(1)/(3)r^3\sin(\theta)\Big]\limits^(6)_0 d\theta \\\\\\\Longrightarrow \int\limits^(\pi/2)_(\pi/4)\Big[((2)/(3)(6)^3\cos(\theta)-(1)/(3)(6)^3\sin(\theta))-(0)\Big] d\theta\\\\\\\Longrightarrow \int\limits^(\pi/2)_(\pi/4)(144\cos(\theta)-72\sin(\theta))d\theta](https://img.qammunity.org/2024/formulas/mathematics/high-school/ekdzrwgol5kvoasxgej5vupk8gptmxvbt9.png)
![\Longrightarrow\Big[144\sin(\theta)+72\cos(\theta)\Big]\limits^(\pi/2)_(\pi/4)\\\\\\\Longrightarrow \Big[(144\sin(\pi/2)+72\cos(\pi/2))-(144\sin(\pi/4)+72\cos(\pi/4))\Big]\\\\\\\Longrightarrow \Big[(144)-(144((√(2) )/(2) )+72((√(2) )/(2) ))\Big] \\ \\ \\ \Longrightarrow \Big[(144)-(72√(2) +36√(2) )\Big]\\\\\\\Longrightarrow \boxed{144-108\sqrt2 \ \text{units}^2}](https://img.qammunity.org/2024/formulas/mathematics/high-school/o4kjwltxi48duxodvksspsw2ie2h4omvmr.png)
Thus, the area of the region is
.