Problem 20 - Calculus Early Transcendentals

Question

Find the mass and centroid (center of mass) of the following thin plates, assuming constant density. Sketch the region corresponding to the plate and indicate the location of the center of mass. Use symmetry when possible to simplify your work. The region bounded by \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=9,\) for \(y \geq 0\)

Step-by-Step Solution

Answer

Question: Find the mass and centroid of the region in the upper half-plane bounded by the circles centered at the origin with radii 1 and 3, assuming constant density. Solution: The mass of the thin plate in the upper half-plane region bounded by \(x^2+y^2=1\) and \(x^2+y^2=9\) with a constant density is \(4\pi\), and the centroid (center of mass) is located at \((0, \frac{8}{3})\).

Step 1: Setting up the Integral for Mass

Since the density is constant, we can calculate the mass of the region by multiplying the area of the region by the density. The area of the region can be found by integrating between the two curves. We can set up the integrals for both the areas enclosed by the two circles. Let A1 be the area of the circle with radius 1, and A2 be the area of the circle with radius 3. In polar coordinates, the equation for the circle is \(r^{2} = x^{2} + y^{2}\). Thus, the equations are \(r^2 = 1\) and \(r^2=9\). So, \(A1 = \int_{0}^{\pi/2}{\int_{0}^{1}{r dr d\theta}}\) and \(A2 = \int_{0}^{\pi/2}{\int_{0}^{3}{r dr d\theta}}\)

Step 2: Calculate the Area

Now we can evaluate the integrals to find the area of the two regions: \(A1 = \int_{0}^{\pi/2}{\int_{0}^{1}{r dr d\theta}} = \left[\frac{r^2}{2}\right]_{0}^{1}\Bigg|_{0}^{\pi/2}= \frac{\pi}{2}\) \(A2 = \int_{0}^{\pi/2}{\int_{0}^{3}{r dr d\theta}} = \left[\frac{r^2}{2}\right]_{0}^{3}\Bigg|_{0}^{\pi/2} = \frac{9\pi}{2}\) Now, let's find the area of the region we need to find the mass: \(Area = A2 - A1 = \frac{9\pi}{2} - \frac{\pi}{2} = 4\pi\)

Step 3: Setup Integral for the Centroid

We need to find the centroid of the region. To do this, we will find the x-coordinate and y-coordinate of the centroid separately. We can integrate the product of the distance from the origin and area with respect to the coordinates. We can write the center of mass coordinates \((\bar{x}, \bar{y})\) as follows: \(\bar{x} = \frac{1}{Area} \int_{R} x dA\) \(\bar{y} = \frac{1}{Area} \int_{R} y dA\) where R is the region we are considering.

Step 4: Calculate Center of Mass Coordinates

In polar coordinates, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Now we can set up the integral for the x and y coordinates of the centroid: \(\bar{x} = \frac{1}{4\pi} \int_{0}^{\pi/2}{\int_{1}^{3}{r^2 \cos(\theta) dr d\theta}}\) \(\bar{y} = \frac{1}{4\pi} \int_{0}^{\pi/2}{\int_{1}^{3}{r^2 \sin(\theta) dr d\theta}}\) Now, we can evaluate the integrals: \(\bar{x} = \frac{1}{4\pi} \int_{0}^{\pi/2}{\int_{1}^{3}{r^2 \cos(\theta) dr d\theta}} = \frac{1}{4\pi} \left[\frac{r^3}{3}\right]_{1}^{3} \Bigg|_{0}^{\pi/2} \cos(\theta)d\theta = 0\) \(\bar{y} = \frac{1}{4\pi} \int_{0}^{\pi/2}{\int_{1}^{3}{r^2 \sin(\theta) dr d\theta}} = \frac{1}{4\pi} \left[\frac{r^3}{3}\right]_{1}^{3} \Bigg|_{0}^{\pi/2} \sin(\theta)d\theta = \frac{8}{3}\) So, the centroid (center of mass) is located at \((\bar{x}, \bar{y}) = (0, \frac{8}{3})\).

Step 5: Conclusion

The mass of the thin plate in the upper half-plane region bounded by \(x^2+y^2=1\) and \(x^2+y^2=9\) with a constant density is \(4\pi\), and the centroid (center of mass) is located at \((0, \frac{8}{3})\).