Determine the area of the region enclosed by the limaçon
r=5+2sin(theta)
You have to integrate 1/2 r^2 from theta = 0 to 2 pi.
1/2 r^2 = 1/2 [25 + 20 sin(theta)
+ 4 sin^2(theta)]
The sin(theta) term doesn't contribute to the integral and sin^2(theta) can be replaced by 1/2 [because it would yield the same contribution as cos^2(theta) and together they are equal to 1].
So, the area is given by:
pi[25 + 2 ] = 27 pi
To determine the area of the region enclosed by the limaçon \(r = 5 + 2\sin(\theta)\), we first need to compute the limits of integration.
The limaçon equation is given in polar coordinates (radius \(r\) as a function of angle \(\theta\)). To find the limits of integration for \(\theta\), we need to identify the angles where the enclosed region starts and ends.
In this case, the limaçon has a petal-shaped loop. The limaçon starts at \(\theta\) with \(\sin(\theta) = 0\) (i.e., \(\theta = 0,\pi\)), and it loops around at \(\theta\) with \(5 + 2\sin(\theta) = 0\) (i.e., \(\theta = \pi,\,2\pi\)).
Therefore, the limits of integration for \(\theta\) are \(\theta = 0\) to \(\theta = 2\pi\).
To find the area of the region using polar coordinates, we can integrate the expression \(1/2 \cdot r^2 d\theta\) over the limits of integration:
\[A = \frac{1}{2} \int_{0}^{2\pi} (5 + 2\sin(\theta))^2 d\theta\]
To evaluate this integral, we expand the expression inside the integral:
\[A = \frac{1}{2} \int_{0}^{2\pi} (25 + 20\sin(\theta) + 4\sin^2(\theta)) d\theta\]
Next, we can distribute the integral across the terms:
\[A = \frac{1}{2} \int_{0}^{2\pi} 25d\theta + \frac{1}{2} \int_{0}^{2\pi} 20\sin(\theta) d\theta + \frac{1}{2} \int_{0}^{2\pi} 4\sin^2(\theta) d\theta\]
Simplifying further:
\[A = \frac{25}{2} \int_{0}^{2\pi} d\theta + 10 \int_{0}^{2\pi} \sin(\theta) d\theta + 2 \int_{0}^{2\pi} \sin^2(\theta) d\theta\]
We know that \(\int \sin(\theta) d\theta = -\cos(\theta)\) and \(\int \sin^2(\theta) d\theta = \frac{\theta}{2} - \frac{\sin(2\theta)}{4}\). Substituting these values:
\[A = \frac{25}{2} (\theta \bigg|_{0}^{2\pi}) + 10 (-\cos(\theta) \bigg|_{0}^{2\pi}) + 2 \left(\frac{\theta}{2} - \frac{\sin(2\theta)}{4} \bigg|_{0}^{2\pi}\right)\]
Evaluating the limits:
\[A = \frac{25}{2} (2\pi - 0) + 10 (-\cos(2\pi) + \cos(0)) + 2 \left(\frac{2\pi}{2} - \frac{\sin(4\pi)}{4} - \left(\frac{0}{2} - \frac{\sin(0)}{4}\right)\right)\]
We know that \(\cos(2\pi) = \cos(0) = 1\) and \(\sin(4\pi) = \sin(0) = 0\), so the expression simplifies to:
\[A = \frac{25}{2} (2\pi) + 10 (-1 + 1) + 2 \left(\pi - 0\right)\]
Simplifying further:
\[A = 25\pi + 2\pi = 27\pi\]
Therefore, the area of the region enclosed by the limaçon \(r = 5 + 2\sin(\theta)\) is \(27\pi\).