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

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

Commented bymr W last updated on 09/Jul/20

you are right. i misread, because this  is how the question is displayed on  my device:

Commented bymr W last updated on 08/Jul/20

A=∫_0 ^(2π) ∫_(4−2cosθ) ^6 rdrdθ

Commented byAr Brandon last updated on 08/Jul/20

Why the 6 at the upper bound, mr W ?

Commented bymr W last updated on 09/Jul/20

Commented bymr W last updated on 09/Jul/20

i misinterpreted as from curve  r=4−2 cos θ to curve r=6.

Commented bybobhans last updated on 09/Jul/20

sir why ∫_0 ^(2π)  ? i think ∫_0 ^π  ?

Commented bybobhans last updated on 09/Jul/20

Commented byAr Brandon last updated on 09/Jul/20

OK Sir

Commented byAr 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 4−2cosθ implies  4−2cosθ<r  and outer the curve 6+2cosθ implies  6+2cosθ>r  Therefore at points of intersection r=r  ⇒4−2cosθ=6+2cosθ ⇒cosθ=((−1)/2)  ⇒θ_1 =(4/3)π , θ_2 =(2/3)π

Commented byAr Brandon last updated on 09/Jul/20

And 4−2cosθ<r<6+2cosθ

Commented bybemath last updated on 09/Jul/20

Commented by1549442205 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

Commented by1549442205 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

Answered by bemath last updated on 09/Jul/20