Question

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.

Answer 1

`144-108\sqrt2 \ \text{units}^2`

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.

`\iint_R (2x-y)dA`

(1) - Changing to polar coordinates from rectangular

`\\\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)((y)/(x))\end{array}\right}`

Limits:

r^2=36 => r=±6

∴ 0≤r≤6

and π/4≤θ≤π/2

Integrand:

`2x-y \rightarrow 2r\cos(\theta)-r\sin(\theta)`

We now have,

`\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 }`

(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`

`\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}`

Thus, the area of the region is
`144-108\sqrt2 \ \text{units}^2`.