find the area bounded inner the curve r = 4−2cos θ and outer the curve r = 6+2cos θ


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Question Number 102369 by bobhans last updated on 08/Jul/20

$f i n d t h e a r e a b o u n d e d i n n e r t h e c u r v e$ $r = 4 - 2 \cos θ a n d o u t e r t h e c u r v e r = 6 + 2 \cos θ$
	Answered by Ar Brandon last updated on 08/Jul/20
	$Area , A=∫_0 ^(2π) ∫_(4−2cosθ) ^(6+2cosθ) rdrdθ=∫_0 ^(2π) [(r^2 /2)]_(4−2cosθ) ^(6+2cosθ) dθ ⇒A=(1/2)∫_0 ^(2π) {(6+2cosθ)^2 −(4−2cosθ)^2 }dθ =(1/2)∫_0 ^(2π) {(36+24cosθ+4cos^2 θ)−(16−16cosθ+4cos^2 θ)}dθ =(1/2)∫_0 ^(2π) (20+40cosθ)dθ=[((20θ+40sinθ)/2)]_0 ^(2π) =20π square units Don′t really know if I′ve done it in the right way. Please let me know what you think.$
	$Area, A = \int_{0}^{2 π} \int_{4 - 2 \cos θ}^{6 + 2 \cos θ} rdrd θ = \int_{0}^{2 π} {[\frac{r^{2}}{2}]}_{4 - 2 \cos θ}^{6 + 2 \cos θ} d θ$ $\Rightarrow A = \frac{1}{2} \int_{0}^{2 π} {{(6 + 2 \cos θ)}^{2} - {(4 - 2 \cos θ)}^{2}} d θ$ $= \frac{1}{2} \int_{0}^{2 π} {(36 + 24 \cos θ + {4 \cos}^{2} θ) - (16 - 16 \cos θ + {4 \cos}^{2} θ)} d θ$ $= \frac{1}{2} \int_{0}^{2 π} (20 + 40 \cos θ) d θ = {[\frac{20 θ + 40 \sin θ}{2}]}_{0}^{2 π}$ $= 20 π square units$ ${D o n}^{'} t r e a l l y k n o w i f I^{'} v e d o n e i t i n t h e r i g h t w a y .$ $P l e a s e l e t m e k n o w w h a t y o u t h i n k .$
	Commented by mr W last updated on 09/Jul/20

	$y o u a r e r i g h t . i m i s r e a d, b e c a u s e t h i s$ $i s h o w t h e q u e s t i o n i s d i s p l a y e d o n$ $m y d e v i c e :$
	Commented by mr W last updated on 08/Jul/20

	$A = \int_{0}^{2 π} \int_{4 - 2 \cos θ}^{6} rdrd θ$
	Commented by Ar Brandon last updated on 08/Jul/20
	Why the 6 at the upper bound, mr W ?
	Commented by mr W last updated on 09/Jul/20

	Commented by mr W last updated on 09/Jul/20

	$i m i s i n t e r p r e t e d a s f r o m c u r v e$ $r = 4 - 2 \cos θ t o c u r v e r = 6 .$
	Commented by bobhans last updated on 09/Jul/20

	$s i r w h y \underset{0}{\int^{2 π}} ? i t h i n k \underset{0}{\int^{π}} ?$
	Commented by bobhans last updated on 09/Jul/20

	Commented by Ar Brandon last updated on 09/Jul/20
	OK Sir
	Commented by Ar Brandon last updated on 09/Jul/20

	${You}^{'} re heading somewhere mr Bobhans . I think {you}^{'} re$ $making a point there . {Let}^{'} s see;$ $Mathematically inner the curve 4 - 2 \cos θ implies$ $4 - 2 \cos θ < r$ $and outer the curve 6 + 2 \cos θ implies$ $6 + 2 \cos θ > r$ $Therefore at points of intersection r = r$ $\Rightarrow 4 - 2 \cos θ = 6 + 2 \cos θ \Rightarrow \cos θ = \frac{- 1}{2}$ $\Rightarrow θ_{1} = \frac{4}{3} π, θ_{2} = \frac{2}{3} π$
	Commented by Ar Brandon last updated on 09/Jul/20

	$And 4 - 2 \cos θ < r < 6 + 2 \cos θ$
	Commented by bemath last updated on 09/Jul/20

	Commented by 1549442205 last updated on 09/Jul/20
	$If the figure is bounded by { ((r=4−2cosθ)),((r=6)) :} then S=∫_0 ^π [6^2 −(4−2cosθ)^2 ]dθ =∫_0 ^π (20+16cosθ−4cos^2 θ)dθ =(20θ+16sinθ)∣_0 ^π −2∫_0 ^π (1+cos2θ)dθ =20𝛑−(2𝛉+sin2𝛉)∣_0 ^𝛑 =18𝛑 If the figure is bounded { ((r=4−2cosθ)),((r=6+2cosθ)) :}then S=∫_0 ^((2π)/3) [(6+2cosθ)^2 −(4−2cosθ)^2 ]dθ =∫_0 ^((2π)/3) (20+40cosθdθ=(20θ+40sinθ)∣_0 ^((2π)/3) =20×((2𝛑)/3)+40×((√3)/2)=((40𝛑)/3)−20(√(3 )) ≈76.53$
	$If the figure is bounded by$ ${\begin{cases} r = 4 - 2 \cos θ \\ r = 6 \end{cases} then$ $S = \int_{0}^{π} [6^{2} - {(4 - 2 \cos θ)}^{2}] d θ$ $= \int_{0}^{π} (20 + 16 \cos θ - {4 \cos}^{2} θ) d θ$ $= (20 θ + 16 \sin θ) ∣_{0}^{π} - 2 \int_{0}^{π} (1 + \cos 2 θ) d θ$ $= 20 π - (2 θ + \sin 2 θ) ∣_{0}^{π} = 18 π$ $If the figure is bounded {\begin{cases} r = 4 - 2 \cos θ \\ r = 6 + 2 \cos θ \end{cases} then$ $S = \int_{0}^{\frac{2 π}{3}} [{(6 + 2 \cos θ)}^{2} - {(4 - 2 \cos θ)}^{2}] d θ$ $= \int_{0}^{\frac{2 π}{3}} (20 + 40 \cos θ d θ = (20 θ + 40 \sin θ) ∣_{0}^{\frac{2 π}{3}}$ $= 20 \times \frac{2 π}{3} + 40 \times \frac{\sqrt{3}}{2} = \frac{40 π}{3} - 20 \sqrt{3} \approx 76.53$
	Commented by 1549442205 last updated on 09/Jul/20


	Answered by Ar Brandon last updated on 09/Jul/20
	$Area , A=∫_((2π)/3) ^((4π)/3) ∫_(4−2cosθ) ^(6+2cosθ) rdrdθ=∫_((2π)/3) ^((4π)/3) [(r^2 /2)]_(4−2cosθ) ^(6+2cosθ) dθ ⇒A=(1/2)∫_((2π)/3) ^((4π)/3) {(6+2cosθ)^2 −(4−2cosθ)^2 }dθ =(1/2)∫_((2π)/3) ^((4π)/3) {(36+24cosθ+4cos^2 θ)−(16−16cosθ+4cos^2 θ)}dθ =(1/2)∫_((2π)/3) ^((4π)/3) (20+40cosθ)dθ=[((20θ+40sinθ)/2)]_((2π)/3) ^((4π)/3) =(1/2)[20×((4π)/3)−40×((√3)/2)−20×((2π)/3)+40×((√3)/2)] =((20)/3)π square units$
	$Area, A = \int_{\frac{2 π}{3}}^{\frac{4 π}{3}} \int_{4 - 2 \cos θ}^{6 + 2 \cos θ} rdrd θ = \int_{\frac{2 π}{3}}^{\frac{4 π}{3}} {[\frac{r^{2}}{2}]}_{4 - 2 \cos θ}^{6 + 2 \cos θ} d θ$ $\Rightarrow A = \frac{1}{2} \int_{\frac{2 π}{3}}^{\frac{4 π}{3}} {{(6 + 2 \cos θ)}^{2} - {(4 - 2 \cos θ)}^{2}} d θ$ $= \frac{1}{2} \int_{\frac{2 π}{3}}^{\frac{4 π}{3}} {(36 + 24 \cos θ + {4 \cos}^{2} θ) - (16 - 16 \cos θ + {4 \cos}^{2} θ)} d θ$ $= \frac{1}{2} \int_{\frac{2 π}{3}}^{\frac{4 π}{3}} (20 + 40 \cos θ) d θ = {[\frac{20 θ + 40 \sin θ}{2}]}_{\frac{2 π}{3}}^{\frac{4 π}{3}}$ $= \frac{1}{2} [20 \times \frac{4 π}{3} - 40 \times \frac{\sqrt{3}}{2} - 20 \times \frac{2 π}{3} + 40 \times \frac{\sqrt{3}}{2}]$ $= \frac{20}{3} π square units$
	Answered by bemath last updated on 09/Jul/20

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