To find the area enclosed by one loop of the curve given in polar coordinates as r = 4 cos(3θ), we will use the polar area formula:
A = \( \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta \)
First, we need to determine the limits of integration, \(\alpha\) and \(\beta\), which correspond to the angles where the loop starts and ends. Since this is a rose curve with 3 petals, we will calculate the area for one petal. The petal of the curve will complete one loop from 0 to \(\frac{\pi}{3}\), where the cosine function returns to its starting point.
Next, we find \(r^2\):
\(r^2 = (4 \cos(3θ))^2 = 16 \cos^2(3θ)\)
Now we can substitute this into our area formula:
\(A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} 16 \cos^2(3θ) dθ\)
We can factor out the constant:
\(A = 8 \int_{0}^{\frac{\pi}{3}} \cos^2(3θ) dθ\)
To integrate \(\cos^2(3θ)\), we can use the identity:
\(\cos^2(x) = \frac{1 + \cos(2x)}{2}\)
So, substituting this into the integral, we have:
\(A = 8 \int_{0}^{\frac{\pi}{3}} \frac{1 + \cos(6θ)}{2} dθ\)
Now the integral becomes:
\(A = 4 \int_{0}^{\frac{\pi}{3}} (1 + \cos(6θ)) dθ\)
This can be evaluated as two separate integrals:
\(A = 4 \left( \int_{0}^{\frac{\pi}{3}} 1 dθ + \int_{0}^{\frac{\pi}{3}} \cos(6θ) dθ \right)\)
The first part evaluates to:
\(\int_{0}^{\frac{\pi}{3}} 1 dθ = \frac{\pi}{3}\)
And the second part evaluates to:
\(\int \cos(6θ) dθ = \frac{1}{6} \sin(6θ)\), so:
\(\int_{0}^{\frac{\pi}{3}} \cos(6θ) dθ = \frac{1}{6} \left( \sin(2\pi) – \sin(0) \right) = 0\)
Combining these results, we find:
\(A = 4 \left( \frac{\pi}{3} + 0 \right) = \frac{4\pi}{3}\)
Thus, the area of the region enclosed by one loop of the curve r = 4 cos(3θ) is:
\(\frac{4\pi}{3}\)