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# Find the area of the inner loop of the limaçon with polar equation r=7cosθ−2?

θ1=arccos(2/7)

### 4 Answers

- AlanLv 79 months agoFavorite Answer
I agree with answer which says

= ∫ [ 0 to arccos(2/7) ] ( 49 ( 1 + cos(2u))/2 + 4 - 28cos(u)) du

( I am changing everything back to theta, since it helps me )

= ∫ [ 0 to arccos(2/7) ] ( 49 ( 1 + cos(2θ))/2 + 4 - 28cos(θ) dθ

but that gives

= [ 0 to arccos(2/7) ] 49*θ/2 + 49*sin(2θ)/4 + 4θ -28*sin(θ)

= [0 to arccos(2/7) ] 57*θ/2 + 49*sin(θ)*cos(θ)/2 -28*sin(θ)

since cos(θ) = 2/7

in Q1

sin(θ) = sqrt( 1 - (2/7)^2) = sqrt( 45/49) = (3/7)*sqrt(5)

= 57*acos(2/7) /2 + 49*(2/7)*(3/7)*sqrt(5)/2 -28*(3/7)*sqrt(5)

- [ 0 + 0 +0 ]

= 57*acos(2/7) /2 + 49*(2/7)*(3/7)*sqrt(5)/2 -28*(3/7)*sqrt(5)

= approx. 16.38516003

- IndikosLv 79 months ago
I use u for theta

The curve starts at u=0 and first comes to origin when

r = 0 or 7cos u - 2 =0

u = arccos(2/7)

So from u = 0 to u=arccos(2/7) it traces out the upper half of the inner loop.

Area of inner loop = 2 * ∫ [ 0 to arccos(2/7) ] 1/2 * r^2 du

= ∫ [ 0 to arccos(2/7) ] ( 7 cos(u) - 2)^2 du

= ∫ [ 0 to arccos(2/7) ] ( 49 cos^2(u) + 4 - 28cos(u)) du

= ∫ [ 0 to arccos(2/7) ] ( 49 ( 1 + cos(2u))/2 + 4 - 28cos(u)) du

= 49( x + sin(2u)/2 ) / 2 + 4x - 28sin(u) [ 0 to arccos(2/7)

= 49( arccos(2/7) + sin(2*arccos(2/7)/2)/2 + 4 arccos(2/7) - 28 sin(arccos(2/7))

- 49( arccos(0) + sin(2*arccos(0)/2)/2 + 4 arccos(0) - 28 sin(arccos(0))

Misplaced or missing parenthesis may confuse

people read the answer:

::::49( arccos(2/7) + sin(2*arccos(2/7)/2)/2 + 4 arccos(2/7) - 28 sin(arccos(2/7))

:::::: Should be

49 ( arccos(2/7) + (sin(2*arccos(2/7)) /2 )/2 + 4 arccos(2/7) - 28 sin(arccos(2/7))- Login to reply the answers

- Ian HLv 79 months ago
Area enclosed by a closed loop given in polar is

A = {θ: θ1 to θ2} ∫(1/2)r^2 dθ

The limaçon r = 7cosθ – 2 self intersects as shown

θ1 = arccos(2/7) ~ 1.281045

The graph and symmetry gives us the useful clue that

θ2 = -arccos(2/7) ~ 2𝛑 - 1.281044625 = 5.002140

A = {θ: θ1 to θ2} ∫{(49/2)[cos(θ)]^2 - 14 cos(θ) + 2} dθ

A =|(49/4)sc – 14s + (57/4)θ| from θ1 to θ2

You should get ~ 73

- ...Show all comments
Your integral is summing thin triangles within the loop. Going from bottom to top, the limits are (-1.281045) to (+1.281045).

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This answer (16.38516003) agrees wolfram answer