Problem 23 - 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 triangular plate in the first quadrant bounded by \(x+y=4\) with \(\rho(x, y)=1+x+y\)

Step-by-Step Solution

Answer

Answer: The center of mass of the triangular region is at point (2,2).

Step 1: Calculate the mass of the region

We will calculate the mass by integrating the density function over the entire region: \(M = \iint_R \rho(x,y) dA\) First we need to find the limits of the integral to define the region: \(0 \leq y \leq 4-x\) \(0 \leq x \leq 4\) The region being a triangle in the first quadrant, we can calculate the double integral as follows: \(M = \int_0^4 \int_0^{4-x} (1+x+y) dydx\)

Step 2: Calculate the moments

Next, we will calculate the moments of the region as follows: 1. Moment about x-axis: \(\bar{y} = \frac{1}{M} \iint_R y \rho(x, y) dA\) 2. Moment about y-axis: \(\bar{x} = \frac{1}{M} \iint_R x \rho(x, y) dA\) Keeping the same limits as before, we can write the double integrals: \(\bar{y}M = \int_0^4 \int_0^{4-x} y(1+x+y) dydx\) \(\bar{x}M = \int_0^4 \int_0^{4-x} x(1+x+y) dydx\)

Step 3: Evaluate the double integrals

Now we will evaluate each double integral one by one: 1. Calculating the mass \(M\): \(M = \int_0^4 \int_0^{4-x} (1+x+y) dydx = \int_0^4 (8 - \frac{1}{2}x^3) dx = 28\) 2. Calculating the moment about x-axis \(\bar{y}M\): \(\bar{y}M = \int_0^4 \int_0^{4-x} y(1+x+y) dydx = \int_0^4 (\frac{8}{3} - \frac{1}{4}x^4) dx = \frac{224}{3}\) 3. Calculating the moment about y-axis \(\bar{x}M\): \(\bar{x}M= \int_0^4 \int_0^{4-x} x(1+x+y) dydx = \int_0^4 (\frac{8}{3} - \frac{1}{4}x^4) dx = \frac{224}{3}\)

Step 4: Calculate the center of mass

Finally, we will calculate the center of mass by dividing the moments by the mass: \(\bar{x} = \frac{\bar{x}M}{M} = \frac{224/3}{28} = 2\) \(\bar{y} = \frac{\bar{y}M}{M} = \frac{224/3}{28} = 2\) Thus, the center of mass of the triangular region is \((\bar{x},\bar{y}) = (2,2)\).