find the area of the region inside the circle r = 3sinè and outside the cardioid r = 1 + sinè
To find the area of the region inside the circle \(r = 3\sin(\theta)\) and outside the cardioid \(r = 1 + \sin(\theta)\), we need to set up and evaluate an integral.
The first step is to identify the range of values for \(\theta\) that define the region of interest. Both equations have \(\theta\)-dependence, so we need to determine where the two curves intersect.
To find the points of intersection, we set \(3\sin(\theta) = 1 + \sin(\theta)\) and solve for \(\theta\):
\(3\sin(\theta) = 1 + \sin(\theta)\)
\(2\sin(\theta) = 1\)
\(\sin(\theta) = \frac{1}{2}\)
From this equation, we get two possible values for \(\theta\): \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
Next, we can set up the integral to calculate the area. We need to integrate the difference between the two curves in polar coordinates, using \(\theta\) as the variable. The formula for the area enclosed by two polar curves is:
\(A = \frac{1}{2}\int_{\alpha}^{\beta}\left[f_1(\theta)^2 - f_2(\theta)^2\right]d\theta\)
Where \(\alpha\) and \(\beta\) are the angles where the curves intersect, and \(f_1(\theta)\) and \(f_2(\theta)\) are the two curves themselves.
Given that \(r_1 = 3\sin(\theta)\) and \(r_2 = 1 + \sin(\theta)\), the integral becomes:
\(A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left[(3\sin(\theta))^2 - (1 + \sin(\theta))^2\right]d\theta\)
Now, we can simplify and evaluate the integral to find the area.