# Area enclosed by polar curve

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.

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.

• 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 $2\pi$? Dec 26 '18 at 20:05

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.

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