$r=(\theta-1)(\theta-3)(\theta-5).$

The diagram above shows a plot of the polar equationFind 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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Recall that the area integral with polar coordinates is

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

First Part - Area of the Large LoopsIf $r = 0$ , $\theta = 1,3,5$ . Notice that $(\theta - 1)(\theta - 3)(\theta - 5)$ forms two closed loops at $1 \leq \theta \leq 3$ and at $3 \leq \theta \leq 5$ . Then,

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

By setting $u = \theta - 3$ , we obtain $u^2(u^2 - 4)^2$ for $-2 \leq u \leq 2$ , which is the even function. In that case, two loops have the same area. From here,

$\begin{array}{rl} A_{\text{two loops}} &= A_{\text{closed loop}_1} + A_{\text{closed loop}_2}\\ &= \int\limits_{u=0}^{u=2} \left(u^6 - 4u^4 - 4u^2 \right) du\\ &= \dfrac{1024}{105} \end{array}$

Second Part - Removing the Small AreaWe 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 $A_{\text{small loop}}$ . We can't find the intersection point of two loops (where $r \neq 0$ ) easily by setting two unique $\theta$ '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 $\theta = 3$ , the shared points of two large loops. Converting to rectangular coordinates, $\theta = 3 \Rightarrow y = \tan(3)x$ The equation of symmetry is $y = -\cot(3)x$ which in polar coordinates becomes $\theta = \arctan(-\cot(3))$ 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 $u$ -substitution, $\begin{array}{rl} A_{\text{small loop}} &= 2A_{\text{halved loop}}\\ &= \int\limits_{-2}^{\arctan(-\cot(3)) - 3} u^2(u^2 - 4)^2 du\\ &\approx1.012 \end{array}$ Thus, the difference between $A_{\text{two loops}}$ and $A_{\text{small loop}}$ is about $\boxed{8.741}$ .