Problem 25 - Calculus Early Transcendentals

Question

Find the center of mass of the following plane regions with variable density. Describe the distribution of mass in the region. The quarter disk in the first quadrant bounded by \(x^{2}+y^{2}=4\) with \(\rho(x, y)=1+x^{2}+y^{2}\)

Step-by-Step Solution

Answer

Answer: The center of mass is at \((x_{cm}, y_{cm}) = \left(\frac{2}{3}, \frac{2}{3}\right)\).

Step 1: Convert the density function into polar coordinates

We are given the density function in Cartesian coordinates: \(\rho(x,y) = 1 + x^2 + y^2\). In polar coordinates, we have \(x = r\cos\theta\) and \(y = r\sin\theta\). So, let's substitute these expressions into the density function to get it in polar coordinates: \(\rho(r,\theta) = 1 + (r\cos\theta)^2 + (r\sin\theta)^2 = 1 + r^2(\cos^2\theta + \sin^2\theta) = 1 + r^2\). Now, the density function is given by \(\rho(r,\theta) = 1 + r^2\).

Step 2: Calculate the total mass

Now let's calculate the total mass \(M\) of the region by integrating the density function over the region in polar coordinates: \(M = \int_R \rho(r, \theta) dA = \int_0^{\frac{\pi}{2}} \int_0^2 (1 + r^2) r dr d\theta\). First, solve the inner integral with respect to \(r\): \(\int_0^2 (1 + r^2) r dr = \int_0^2 (r + r^3) dr = \left[\frac{1}{2}r^2 + \frac{1}{4}r^4\right]_0^2 = (\frac{1}{2}(4) + \frac{1}{4}(16)) = 4 + 4 = 8\). Now, solve the outer integral: \(\int_0^{\frac{\pi}{2}} 8 d\theta = 8\left[\theta\right]_0^{\frac{\pi}{2}} = 8\left(\frac{\pi}{2}\right) = 4\pi\). So, the total mass \(M\) of the region is \(4\pi\).

Step 3: Calculate \(x_{cm}\) and \(y_{cm}\)

Now, let's calculate the coordinates of the center of mass using the formulas \(x_{cm} = \frac{1}{M} \int_R x \rho(x, y) dA\) and \(y_{cm} = \frac{1}{M} \int_R y \rho(x, y) dA\). In polar coordinates, this translates to: \(x_{cm} = \frac{1}{4\pi} \int_R r\cos\theta \rho(r, \theta) r dr d\theta\) and \(y_{cm} = \frac{1}{4\pi} \int_R r\sin\theta \rho(r, \theta) r dr d\theta\). First, calculate \(x_{cm}\): \(x_{cm} = \frac{1}{4\pi} \int_0^{\frac{\pi}{2}} \int_0^2 r\cos\theta (1+r^2) r dr d\theta = \frac{1}{4\pi} \int_0^{\frac{\pi}{2}} \int_0^2 r^2\cos\theta (1+r^2) dr d\theta\). Solve the inner integral: \(\int_0^2 r^2\cos\theta (1+r^2) dr = \cos\theta\int_0^2 (r^2 + r^4) dr = \cos\theta\left[\frac{1}{3}r^3 + \frac{1}{5}r^5\right]_0^2 = \frac{8}{3}\cos\theta\). Solve the outer integral: \(\int_0^{\frac{\pi}{2}} \frac{8}{3}\cos\theta d\theta = \frac{8}{3}\left[\sin\theta\right]_0^{\frac{\pi}{2}} = \frac{8}{3}\). Now, multiply by the mass: \(x_{cm} = \frac{1}{4\pi}\cdot\frac{8}{3} = \frac{2}{3}\). Next, calculate \(y_{cm}\): \(y_{cm} = \frac{1}{4\pi} \int_0^{\frac{\pi}{2}} \int_0^2 r\sin\theta (1+r^2) r dr d\theta\). Solve the inner integral: \(\int_0^2 r^2\sin\theta (1+r^2) dr = \sin\theta\int_0^2 (r^2 + r^4) dr = \sin\theta\left[\frac{1}{3}r^3 + \frac{1}{5}r^5\right]_0^2 = \frac{8}{3}\sin\theta\). Solve the outer integral: \(\int_0^{\frac{\pi}{2}} \frac{8}{3}\sin\theta d\theta = \frac{8}{3}\left[-\cos\theta\right]_0^{\frac{\pi}{2}}= \frac{8}{3}\). Now, multiply by the mass: \(y_{cm} = \frac{1}{4\pi}\cdot\frac{8}{3} = \frac{2}{3}\). So, the center of mass is at \((x_{cm}, y_{cm}) = \left(\frac{2}{3}, \frac{2}{3}\right)\).