Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=1+\cos \theta\)

Short answer

Answer: The centroid of the region is located at the point \((0, \frac{5}{12})\).

Step-by-step solution

Identify the region

First, let's identify the region bounded by the cardioid \(r = 1 + \cos \theta\). This curve is traced out in the polar coordinate plane as \(\theta\) ranges from \(0\) to \(2\pi\).

Choose the differential element

Choose a small differential element in the region given by the cardioid. In this case, the differential element is given by \(dA = r\ dr\ d\theta\).

Determine the bounds of r and \(\theta\)

We need to determine the bounds of \(r\) and \(\theta\) for which the differential element lies within the region bounded by the cardioid. Tracing the cardioid, we find that \(\theta\) ranges from \(0\) to \(2\pi\). The cardioid \(r = 1 + \cos \theta\) defines the boundary, so the bounds of \(r\) go from \(0\) to \(1 + \cos \theta\).

Find the area of the region

To find the area of the region, we integrate the differential area element over the entire region. The area is given by: \(A = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r\ dr\ d\theta\)

Find the x-coordinate of the centroid

To find the x-coordinate of the centroid, we need to find the moment of the region about the y-axis and divide it by the area. The x-coordinate of a point in polar coordinates is given by \(x = r\cos \theta\). The differential moment about the y-axis is \(x\ dA = r^2\cos \theta\ dr\ d\theta\). Thus, the moment of the region about the y-axis is: \(M_x=\int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r^2\cos \theta\ dr\ d\theta\) The x-coordinate of the centroid is: \(\bar{x} = \frac{M_x}{A}\)

Find the y-coordinate of the centroid

To find the y-coordinate of the centroid, we need to find the moment of the region about the x-axis and divide it by the area. The y-coordinate of a point in polar coordinates is given by \(y = r\sin \theta\). The differential moment about the x-axis is \(y\ dA = r^2\sin \theta\ dr\ d\theta\). Thus, the moment of the region about the x-axis is: \(M_y=\int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r^2\sin \theta\ dr\ d\theta\) The y-coordinate of the centroid is: \(\bar{y} = \frac{M_y}{A}\)

Evaluate the integrals

Evaluate each of the integrals to find \(A, M_x, M_y\). Then compute \(\bar{x}\) and \(\bar{y}\) using the formulas found in steps 5 and 6. Area: \(A = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r\ dr\ d\theta = \frac{3\pi}{2}\) Moment about \(x\)-axis: \(M_x=\int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r^2\cos \theta\ dr\ d\theta= 0\) Moment about \(y\)-axis: \(M_y=\int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r^2\sin \theta\ dr\ d\theta = \frac{5\pi}{8}\) And the centroid can be calculated as: \(\bar{x} = \frac{M_x}{A} = 0\), \(\bar{y} = \frac{M_y}{A} = \frac{5}{12}\) Therefore, the centroid of the region bounded by the cardioid \(r = 1 + \cos\theta\) is located at the point \((0, \frac{5}{12})\).