Problem 26 - 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 of the plane region is at the point \((x_c, y_c) = \left(\frac{5}{7}, \frac{5}{7}\right)\).

Step 1: Find the total mass M

To find the total mass M, we will integrate the density function over the quarter disk domain: $$M = \iint_{D} (1 + x^2 + y^2) \, dA.$$ Since it is a polar coordinate problem, we can convert the integral to polar coordinates. The relation between Cartesian coordinates and polar coordinates is \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Hence, the density function will be: $$\rho(r, \theta) = 1 + r^2 \cos^2(\theta) + r^2 \sin^2(\theta),$$ and \(dA = r \, dr \, d\theta\). The limits of integration for \(r\) will be from 0 to 2 (radius of the disk), and for \(\theta\) will be from 0 to \(\frac{\pi}{2}\) (first quadrant). So, $$M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (1 + r^2 (\cos^2(\theta) + \sin^2(\theta))) r \, dr \, d\theta.$$ Now we can calculate the mass M.

Step 2: Calculate Mass M

Integrating the inner integral w.r.t r first: $$M = \int_{0}^{\frac{\pi}{2}} \left[ \frac{1}{2}r^2 + \frac{1}{4}r^4 \right]_{0}^{2} \, d\theta = \int_{0}^{\frac{\pi}{2}} (2 + 4) \, d\theta$$ Now integrating the outer integral w.r.t \(\theta\): $$M = \left[ 6\theta \right]_{0}^{\frac{\pi}{2}} = 3\pi$$ So the total mass of the region is \(M = 3\pi\).

Step 3: Compute x_c and y_c

Now that we have the mass M, we can compute \(x_c\) and \(y_c\). Using polar coordinates, the formula for \(x_c\) and \(y_c\) become: $$x_c = \frac{1}{M} \iint_{D} r \cos(\theta) \rho(r, \theta) r \, dr \, d\theta$$ $$y_c = \frac{1}{M} \iint_{D} r \sin(\theta) \rho(r, \theta) r \, dr \, d\theta$$ With the density function already found in step 1, we can solve these integrals.

Step 4: Calculate x_c and y_c

Computing x_c, we get: $$x_c = \frac{1}{3\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \cos(\theta)(1 + r^2) \, dr \, d\theta.$$ Now, we integrate w.r.t r first: $$x_c = \frac{1}{3\pi} \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{4}r^4 + \frac{1}{7}r^7\right]_{0}^{2} \cos(\theta) \, d\theta$$ After evaluating the antiderivative, and calculating the integral w.r.t \(\theta\), we find: $$x_c = \frac{1}{3\pi} \left[\frac{15\pi}{7}\right] = \frac{5}{7}$$ Computing y_c, we get: $$y_c = \frac{1}{3\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \sin(\theta)(1 + r^2) \, dr \, d\theta.$$ We follow the same procedure as for x_c: $$y_c = \frac{1}{3\pi} \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{4}r^4 + \frac{1}{7}r^7\right]_{0}^{2} \sin(\theta) \, d\theta$$ After evaluating the antiderivative, and calculating the integral w.r.t \(\theta\), we find: $$y_c = \frac{1}{3\pi} \left[\frac{15\pi}{7}\right] = \frac{5}{7}$$

Step 5: Present the final result

The center of mass of the plane region with variable density \(\rho(x, y) = 1 + x^2 + y^2\) bounded by the quarter disk in the first quadrant is: $$(x_c, y_c) = \left(\frac{5}{7}, \frac{5}{7}\right)$$