The diagram above shows a plot of the polar equation
Find the total area of the regions bounded by the curve.
Submit your answer in 3 decimal places.
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Recall that the area integral with polar coordinates is
A = 2 1 α ∫ β ( r ( θ ) ) 2 d θ
First Part - Area of the Large Loops
If r = 0 , θ = 1 , 3 , 5 . Notice that ( θ − 1 ) ( θ − 3 ) ( θ − 5 ) forms two closed loops at 1 ≤ θ ≤ 3 and at 3 ≤ θ ≤ 5 . Then,
A closed loop 1 = 2 1 1 ∫ 3 [ ( θ − 1 ) ( θ − 3 ) ( θ − 5 ) ] 2 d θ and A closed loop 2 = 2 1 3 ∫ 5 [ ( θ − 1 ) ( θ − 3 ) ( θ − 5 ) ] 2 d θ
By setting u = θ − 3 , we obtain u 2 ( u 2 − 4 ) 2 for − 2 ≤ u ≤ 2 , which is the even function. In that case, two loops have the same area. From here,
A two loops = A closed loop 1 + A closed loop 2 = u = 0 ∫ u = 2 ( u 6 − 4 u 4 − 4 u 2 ) d u = 1 0 5 1 0 2 4
Second Part - Removing the Small Area
We are not done with the problem since r forms the small loop between two large loops, which doubles the area of the small loop, call A small loop . We can't find the intersection point of two loops (where r = 0 ) easily by setting two unique θ 's. However, since the curve is symmetric with respect to a straight line, we can determine the linear equation that is perpendicular to the curve at θ = 3 , the shared points of two large loops. Converting to rectangular coordinates, θ = 3 ⇒ y = tan ( 3 ) x The equation of symmetry is y = − cot ( 3 ) x which in polar coordinates becomes θ = arctan ( − cot ( 3 ) ) Drawing the line, we notice that (1) it intersects both endpoints of the small loop, and (2) that it splits the small loops in half (symmetry!). Using the same u -substitution, A small loop = 2 A halved loop = − 2 ∫ arctan ( − cot ( 3 ) ) − 3 u 2 ( u 2 − 4 ) 2 d u ≈ 1 . 0 1 2 Thus, the difference between A two loops and A small loop is about 8 . 7 4 1 .