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I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer. Can you assist in clarification?

Find the area of the region enclosed by $r=4cos(3 \theta)$.

I use $ \frac 12 \int_0^{2\pi} (16cos^2(3\theta) d\theta$. For $cos^2(3\theta)$ I use the identity $\frac12[1+cos(6\theta)]$

This gives me $\frac{16}{4} \int_0^{2\pi} 1+cos(6\theta) d\theta$.

This gives me $4[\int_0^{2\pi}1 d\theta +\frac16\int_0^{12\pi} cos (u) du]$.

The integral of the cosine term is $0$, so I get $\theta $ evaluated from $0$ to $2\pi$. This gives me $4(2\pi)=8\pi$.

When I use a symmetrical method A=$6\int_0^{\pi/6}\frac12(16 cos^2(3\theta)d\theta$ I get $4\pi$. This is the text answer.

Don't understand why my 2 answers don't match.

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Your first answer is twice the correct answer for the following reason: if you let $\theta$ range from $\theta=0$ to $\theta=2\pi$, the curve $r=4\cos(3\theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $\theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $\theta=\pi$; and then, from $\theta=\pi$ to $\theta=2\pi$ you retrace it once more.

In the second method, you find the area of a half of one petal, which you correctly determined to range from $\theta=0$ to $\theta=\dfrac{\pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $\theta=0$ to $\theta=\dfrac{\pi}{6}$ will produce the angle six times as wide, i.e. from $\theta=0$ to $\theta=\pi$, consistent with my explanation above.

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  • $\begingroup$ Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $\pi$ instead of $0$ to $2\pi$? $\endgroup$
    – user163862
    Dec 26 '18 at 20:05
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To provide yet another approach, you can use Green's theorem on the differential form $x \,dy$ to transform the area integral over one petal into a line integral along the petal:

$$\frac13 A = \text{area of petal} = \int\limits_{\text{petal}}\,1\,dA = \int\limits_{\text{boundary of petal}} x\,dy $$

Since $r = 4\cos(3\theta)$, the petal can be parameterized by $$(x(\theta), y(\theta)) = (r(\theta)\cos\theta, r(\theta)\sin(\theta)) = (4\cos\theta\cos(3\theta), 4\sin\theta\cos(3\theta))$$

for $\theta \in \left[-\frac{\pi}6, \frac{\pi}6\right]$.

Differentiating gives $dy = (4\cos\theta\cos(3\theta) - 12\sin\theta\sin(3\theta))\,d\theta$ so

$$\frac13 A = \int_{-\frac{\pi}6}^{\frac{\pi}6}4\cos\theta\cos(3\theta) (4\cos\theta\cos(3\theta) - 12\sin\theta\sin(3\theta))\,d\theta = \frac{4\pi}3$$

so $A = 4\pi$.

The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.

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  • $\begingroup$ ... and which is the wrong answer of the two. $\endgroup$
    – zipirovich
    Dec 26 '18 at 18:51
  • $\begingroup$ @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result. $\endgroup$ Dec 26 '18 at 19:36

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