Dtyjlui

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

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

This gives me $4[int_0^{2pi}1 dtheta +frac16int_0^{12pi} cos (u) du]$.

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

When I use a symmetrical method A=$6int_0^{pi/6}frac12(16 cos^2(3theta)dtheta$ I get $4pi$. 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 $theta=2pi$, the curve $r=4cos(3theta)$ — 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 $theta=2pi$ 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.

$$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$

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:

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

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

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

Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so

$$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$

so $A = 4pi$.

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