91Ó°ÊÓ

The equation of the cardioid is given by \(r=4+4\sin\theta\). To sketch the cardioid, we can first find the important points, i.e., the maximum and minimum values of \(r\). We know that \(\sin\theta\) oscillates between -1 and 1, inclusive. Thus, the minimum value of \(r\) is \(4-4=0\) and the maximum value is \(4+4=8\). The cardioid will start and end at \((r,\theta)=(0,0)\), and its largest value of \(r\) occurs when \(\theta=\pi/2\). So, the cardioid will have a loop-like shape with the loop centered at \((r,\theta)=(0,\pi/2)\). Note: The sketch will be crucial to visually understanding this problem but cannot be provided here.

The bounding curves of the region inside the cardioid are given by the cardioid itself, which is \(r=4+4\sin\theta\). To find the limits of integration for \(\theta\), we need to know where the cardioid begins and ends. From our sketch, we can see that the cardioid starts at \(\theta = 0\) and closes at the origin once again as \(\theta\) completes a full rotation from \(0\) to \(2\pi\). Therefore, the limits of integration for \(\theta\) are \(0\) and \(2\pi\).

Now that we have identified the limits of integration for \(\theta\), we can use the formula for the area of a region in polar coordinates: Area = \(\frac{1}{2}\int_{0}^{2\pi} r^2 d\theta\) Plugging in our cardioid equation: Area = \(\frac{1}{2}\int_{0}^{2\pi} (4+4\sin\theta)^2 d\theta\) Expanding the squared term: Area = \(\frac{1}{2}\int_{0}^{2\pi} (16 + 32\sin\theta + 16\sin^2\theta) d\theta\)

We can now evaluate the integral: Area = \(\frac{1}{2}\int_{0}^{2\pi} (16 + 32\sin\theta + 16\sin^2\theta) d\theta\) This integral can be split into three parts: Area = \(\frac{1}{2}\bigg[\int_{0}^{2\pi} 16 d\theta + \int_{0}^{2\pi} 32\sin\theta d\theta + \int_{0}^{2\pi} 16\sin^2\theta d\theta\bigg]\) Evaluating the first integral: \(\int_{0}^{2\pi} 16 d\theta = 16\theta\big|_{0}^{2\pi} = 32\pi\) Evaluating the second integral: \(\int_{0}^{2\pi} 32\sin\theta d\theta = -32\cos\theta\big|_{0}^{2\pi} = 0\) Evaluating the third integral (using double-angle identity \(\sin^2\theta = \frac{1}{2}(1-\cos 2\theta)\)): \(\int_{0}^{2\pi} 16\sin^2\theta d\theta = 16\int_{0}^{2\pi} \frac{1}{2}(1-\cos 2\theta) d\theta = 8\int_{0}^{2\pi} (1-\cos 2\theta) d\theta\) The new integral can be split into two parts: \(8\int_{0}^{2\pi} (1-\cos 2\theta) d\theta = 8\bigg[\int_{0}^{2\pi} d\theta - \int_{0}^{2\pi} \cos 2\theta d\theta\bigg]\) Evaluating the two integrals: \(8\int_{0}^{2\pi} d\theta = 8\theta\big|_{0}^{2\pi} = 16\pi\) \(8\int_{0}^{2\pi} \cos 2\theta d\theta = 4\sin 2\theta\big|_{0}^{2\pi} = 0\) Now, combining the results: Area = \(\frac{1}{2}(32\pi + 0 + 16\pi) = 24\pi\) Conclusion: The area of the region inside the cardioid \(r=4+4\sin\theta\) is \(24\pi\).