Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$${find}\:{the}\:{area}\:{bounded}\:{inner}\:{the}\:{curve} \\ $$$${r}\:=\:\mathrm{4}−\mathrm{2cos}\:\theta\:{and}\:{outer}\:{the}\:{curve}\:{r}\:=\:\mathrm{6}+\mathrm{2cos}\:\theta \\ $$

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.

$$\mathrm{Area}\:,\:\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{4}−\mathrm{2cos}\theta} ^{\mathrm{6}+\mathrm{2cos}\theta} \mathrm{rdrd}\theta=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left[\frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{4}−\mathrm{2cos}\theta} ^{\mathrm{6}+\mathrm{2cos}\theta} \mathrm{d}\theta \\ $$$$\Rightarrow\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left\{\left(\mathrm{6}+\mathrm{2cos}\theta\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2cos}\theta\right)^{\mathrm{2}} \right\}\mathrm{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left\{\left(\mathrm{36}+\mathrm{24cos}\theta+\mathrm{4cos}^{\mathrm{2}} \theta\right)−\left(\mathrm{16}−\mathrm{16cos}\theta+\mathrm{4cos}^{\mathrm{2}} \theta\right)\right\}\mathrm{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{20}+\mathrm{40cos}\theta\right)\mathrm{d}\theta=\left[\frac{\mathrm{20}\theta+\mathrm{40sin}\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{20}\pi\:\mathrm{square}\:\mathrm{units} \\ $$$$ \\ $$$${Don}'{t}\:{really}\:{know}\:{if}\:{I}'{ve}\:{done}\:{it}\:{in}\:{the}\:{right}\:{way}. \\ $$$${Please}\:{let}\:{me}\:{know}\:{what}\:{you}\:{think}. \\ $$

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

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

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

Commented by mr W last updated on 08/Jul/20

A=∫_0 ^(2π) ∫_(4−2cosθ) ^6 rdrdθ

$$\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{4}−\mathrm{2cos}\theta} ^{\mathrm{6}} \mathrm{rdrd}\theta \\ $$

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

$${i}\:{misinterpreted}\:{as}\:{from}\:{curve} \\ $$$${r}=\mathrm{4}−\mathrm{2}\:\mathrm{cos}\:\theta\:{to}\:{curve}\:{r}=\mathrm{6}. \\ $$

Commented by bobhans last updated on 09/Jul/20

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

$${sir}\:{why}\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:?\:{i}\:{think}\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:? \\ $$

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 4−2cosθ 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

$$\mathrm{You}'\mathrm{re}\:\mathrm{heading}\:\mathrm{somewhere}\:\mathrm{mr}\:\mathrm{Bobhans}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}'\mathrm{re} \\ $$$$\mathrm{making}\:\mathrm{a}\:\mathrm{point}\:\mathrm{there}.\:\mathrm{Let}'\mathrm{s}\:\mathrm{see}; \\ $$$$\mathrm{Mathematically}\:\mathrm{inner}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{4}−\mathrm{2cos}\theta\:\mathrm{implies} \\ $$$$\mathrm{4}−\mathrm{2cos}\theta<\mathrm{r} \\ $$$$\mathrm{and}\:\mathrm{outer}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{6}+\mathrm{2cos}\theta\:\mathrm{implies} \\ $$$$\mathrm{6}+\mathrm{2cos}\theta>\mathrm{r} \\ $$$$\mathrm{Therefore}\:\mathrm{at}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{r}=\mathrm{r} \\ $$$$\Rightarrow\mathrm{4}−\mathrm{2cos}\theta=\mathrm{6}+\mathrm{2cos}\theta\:\Rightarrow\mathrm{cos}\theta=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}\pi\:,\:\theta_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\pi \\ $$

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

And 4−2cosθ<r<6+2cosθ

$$\mathrm{And}\:\mathrm{4}−\mathrm{2cos}\theta<\mathrm{r}<\mathrm{6}+\mathrm{2cos}\theta \\ $$

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

$$\mathrm{If}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{is}\:\mathrm{bounded}\:\mathrm{by} \\ $$$$\begin{cases}{\mathrm{r}=\mathrm{4}−\mathrm{2cos}\theta}\\{\mathrm{r}=\mathrm{6}}\end{cases}\:\:\mathrm{then} \\ $$$$\mathrm{S}=\int_{\mathrm{0}} ^{\pi} \left[\mathrm{6}^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2cos}\theta\right)^{\mathrm{2}} \right]\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left(\mathrm{20}+\mathrm{16cos}\theta−\mathrm{4cos}^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$=\left(\mathrm{20}\theta+\mathrm{16sin}\theta\right)\mid_{\mathrm{0}} ^{\pi} −\mathrm{2}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+\mathrm{cos2}\theta\right)\mathrm{d}\theta \\ $$$$=\mathrm{20}\boldsymbol{\pi}−\left(\mathrm{2}\boldsymbol{\theta}+\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\theta}\right)\mid_{\mathrm{0}} ^{\boldsymbol{\pi}} =\mathrm{18}\boldsymbol{\pi} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{is}\:\mathrm{bounded}\:\begin{cases}{\mathrm{r}=\mathrm{4}−\mathrm{2cos}\theta}\\{\mathrm{r}=\mathrm{6}+\mathrm{2cos}\theta}\end{cases}\mathrm{then} \\ $$$$\mathrm{S}=\int_{\mathrm{0}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \left[\left(\mathrm{6}+\mathrm{2cos}\theta\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2cos}\theta\right)^{\mathrm{2}} \right]\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \left(\mathrm{20}+\mathrm{40cos}\theta\mathrm{d}\theta=\left(\mathrm{20}\theta+\mathrm{40sin}\theta\right)\mid_{\mathrm{0}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \right. \\ $$$$=\mathrm{20}×\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{40}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{40}\boldsymbol{\pi}}{\mathrm{3}}−\mathrm{20}\sqrt{\mathrm{3}\:}\:\approx\mathrm{76}.\mathrm{53} \\ $$

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

$$ \\ $$$$\mathrm{Area}\:,\:\mathrm{A}=\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \int_{\mathrm{4}−\mathrm{2cos}\theta} ^{\mathrm{6}+\mathrm{2cos}\theta} \mathrm{rdrd}\theta=\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \left[\frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{4}−\mathrm{2cos}\theta} ^{\mathrm{6}+\mathrm{2cos}\theta} \mathrm{d}\theta \\ $$$$\Rightarrow\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \left\{\left(\mathrm{6}+\mathrm{2cos}\theta\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2cos}\theta\right)^{\mathrm{2}} \right\}\mathrm{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \left\{\left(\mathrm{36}+\mathrm{24cos}\theta+\mathrm{4cos}^{\mathrm{2}} \theta\right)−\left(\mathrm{16}−\mathrm{16cos}\theta+\mathrm{4cos}^{\mathrm{2}} \theta\right)\right\}\mathrm{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \left(\mathrm{20}+\mathrm{40cos}\theta\right)\mathrm{d}\theta=\left[\frac{\mathrm{20}\theta+\mathrm{40sin}\theta}{\mathrm{2}}\right]_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{4}\pi}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{20}×\frac{\mathrm{4}\pi}{\mathrm{3}}−\mathrm{40}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{20}×\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{40}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{20}}{\mathrm{3}}\pi\:\mathrm{square}\:\mathrm{units} \\ $$

Answered by bemath last updated on 09/Jul/20

Terms of Service

Privacy Policy

Contact: info@tinkutara.com