evaluate the double integral ∬r(2x−y)da, where r is the region in the first quadrant enclosed by the circle x2 y2=36 and the lines x=0 and y=x, by changing to polar coordinates.

[tex]\\\boxed{\left\begin{array}{ccc}\text{Using the following conversions:}\\\\x=r\cos \theta\\\\y=r \sin\theta\\\\r^2=x^2+y^2\\\\\theta=\tan^{-1}(\frac{y}{x})\end{array}\right}[/tex]

Limits:

r^2=36 => r=±6

∴ 0≤r≤6

and π/4≤θ≤π/2

Integrand:

[tex]2x-y \rightarrow 2r\cos(\theta)-r\sin(\theta)[/tex]

We now have,

[tex]\int\limits^{\pi/2}_{\pi/4} \int\limits^6_0{( 2r\cos(\theta)-r\sin(\theta))} \, rdrd\theta \\\\\Longrightarrow \boxed{\int\limits^{\pi/2}_{\pi/4} \int\limits^6_0{( 2r^2\cos(\theta)-r^2\sin(\theta))} \, drd\theta }[/tex]

(2) - Evaluating the integral

[tex]\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[\frac{2}{3}r^3\cos(\theta)-\frac{1}{3}r^3\sin(\theta)\Big]\limits^{6}_0 d\theta \\\\\\\Longrightarrow \int\limits^{\pi/2}_{\pi/4}\Big[(\frac{2}{3}(6)^3\cos(\theta)-\frac{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[/tex]

[tex]\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(\frac{\sqrt{2} }{2} )+72(\frac{\sqrt{2} }{2} ))\Big] \\ \\ \\ \Longrightarrow \Big[(144)-(72\sqrt{2} +36\sqrt{2} )\Big]\\\\\\\Longrightarrow \boxed{144-108\sqrt2 \ \text{units}^2}[/tex]

Thus, the area of the region is [tex]144-108\sqrt2 \ \text{units}^2[/tex].