A gift for myself (330 followers problem-3)

Calculus Level pending

The diagram above shows a plot of the polar equation r = ( θ 1 ) ( θ 3 ) ( θ 5 ) . r=(\theta-1)(\theta-3)(\theta-5).

Find the total area of the regions bounded by the curve.

Submit your answer in 3 decimal places.


The answer is 8.74073.

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1 solution

Michael Huang
Oct 31, 2016

Recall that the area integral with polar coordinates is

A = 1 2 α β ( r ( θ ) ) 2 d θ A = \dfrac{1}{2} \int\limits_{\alpha}^{\beta} \left(r(\theta)\right)^2 d\theta

First Part - Area of the Large Loops

If r = 0 r = 0 , θ = 1 , 3 , 5 \theta = 1,3,5 . Notice that ( θ 1 ) ( θ 3 ) ( θ 5 ) (\theta - 1)(\theta - 3)(\theta - 5) forms two closed loops at 1 θ 3 1 \leq \theta \leq 3 and at 3 θ 5 3 \leq \theta \leq 5 . Then,

A closed loop 1 = 1 2 1 3 [ ( θ 1 ) ( θ 3 ) ( θ 5 ) ] 2 d θ and A closed loop 2 = 1 2 3 5 [ ( θ 1 ) ( θ 3 ) ( θ 5 ) ] 2 d θ A_{\text{closed loop}_1} = \dfrac{1}{2} \int\limits_{1}^3 \left[(\theta - 1)(\theta - 3)(\theta - 5)\right]^2 d\theta \qquad \text{and} \qquad A_{\text{closed loop}_2} = \dfrac{1}{2} \int\limits_{3}^5 \left[(\theta - 1)(\theta - 3)(\theta - 5)\right]^2 d\theta

By setting u = θ 3 u = \theta - 3 , we obtain u 2 ( u 2 4 ) 2 u^2(u^2 - 4)^2 for 2 u 2 -2 \leq u \leq 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 = 1024 105 \begin{array}{rl} A_{\text{two loops}} &= A_{\text{closed loop}_1} + A_{\text{closed loop}_2}\\ &= \int\limits_{u=0}^{u=2} \left(u^6 - 4u^4 - 4u^2 \right) du\\ &= \dfrac{1024}{105} \end{array}

Second Part - Removing the Small Area

We are not done with the problem since r r forms the small loop between two large loops, which doubles the area of the small loop, call A small loop A_{\text{small loop}} . We can't find the intersection point of two loops (where r 0 r \neq 0 ) easily by setting two unique θ \theta '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 \theta = 3 , the shared points of two large loops. Converting to rectangular coordinates, θ = 3 y = tan ( 3 ) x \theta = 3 \Rightarrow y = \tan(3)x The equation of symmetry is y = cot ( 3 ) x y = -\cot(3)x which in polar coordinates becomes θ = arctan ( cot ( 3 ) ) \theta = \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 u -substitution, A small loop = 2 A halved loop = 2 arctan ( cot ( 3 ) ) 3 u 2 ( u 2 4 ) 2 d u 1.012 \begin{array}{rl} A_{\text{small loop}} &= 2A_{\text{halved loop}}\\ &= \int\limits_{-2}^{\arctan(-\cot(3)) - 3} u^2(u^2 - 4)^2 du\\ &\approx1.012 \end{array} Thus, the difference between A two loops A_{\text{two loops}} and A small loop A_{\text{small loop}} is about 8.741 \boxed{8.741} .

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