Question

The correct answer is letter E, but I am not sure how to approach this problem and I was wondering if anyone can offer any guidance.

Answer 1

Explanation:

The key to area in polar coordinates is the formula for the area of a sector:

  a = (1/2)r²θ

Then a differential of area* can be written as ...

  da = (1/2)r²·dθ

Filling in the given function for r, we have ...

  da = (1/2)(4cos(3θ))²·dθ = 8cos(3θ)²·dθ

The integral will have limits corresponding to the range of values of θ for one loop of the graph: -π/6 to π/6. So, the area is ...

  $\displaystyle A=\int{da}=\int\limits_{-\pi /6}^{\pi /6}{8\cos{(3\theta)}^2}\, d\theta\\\\=8\int\limits_{-\pi /6}^{\pi /6}{\cos{(3\theta)}^2}$

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* As with other approaches to finding area (horizontal or vertical slice, for example), we assume that the differential element dθ is sufficiently small that we need not concern ourselves with the fact that r is a function of θ.