Problem 18 - 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 \(y=e^{x}, y=e^{-x}, x=0,\) and \(x=\ln 2\)

Step-by-Step Solution

Answer

Based on the solution and given information, the mass of the thin plate with constant density is \(M = \rho(1-\ln 2)\), and the centroid (center of mass) coordinates are \((\bar{x},\bar{y}) = (\ln 2, \frac{1}{4}(e^2 - e^{-2}))\).

Step 1: Sketch the region

First, let's sketch the region bounded by the given curves \(y=e^x, y=e^{-x}, x=0,\) and \(x=\ln 2\). These curves are the graphs of exponential functions, and we know that \(e^x\) and \(e^{-x}\) will intersect at \(x=0\) and \(x=\ln 2\). So, we can sketch the region and identify it as an area bounded by these curves within the given limits.

Step 2: Calculate the mass of the thin plate

Now that we have the region sketched, let us calculate the mass (\(M\)) of the thin plate by finding the double integral of the density function (\(\rho\)) with respect to the area. Since the density is constant, we can just multiply the density by the area of the region. To find the area, we need to perform a double integral of the function \(1\) over the region. Using polar coordinates, we have: \(M = \rho \int_{0}^{\ln 2}\int_{e^x}^{e^{-x}} dy dx\)

Step 3: Find the centroid coordinates (x,y)

Next, we need to find the centroid (center of mass) coordinates \((\bar{x},\bar{y})\) of the thin plate. To do this, we will calculate the double integral of the coordinate functions multiplied by the density function over the region and then divide these by the mass: \(\bar{x} = \frac{1}{M}\int_{0}^{\ln 2}\int_{e^x}^{e^{-x}} x dy dx\) \(\bar{y} = \frac{1}{M}\int_{0}^{\ln 2}\int_{e^x}^{e^{-x}} y dy dx\)

Step 4: Calculate the mass and centroid coordinates

Now, let's compute the actual values for the mass and centroid coordinates using the double integrals: Mass, \(M = \rho \int_{0}^{\ln 2}(e^{-x} - e^x) dx = \rho [(2-1)-(\ln 2)(e^{\ln 2}-1)]=\rho(1-(\ln 2))(1)\) \(\bar{x} = \frac{1}{M}\int_{0}^{\ln 2} x (e^{-x} - e^x) dx = \frac{1}{M}(2(\ln 2)-\ln 2)=(\ln2)\) \(\bar{y} = \frac{1}{M}\int_{0}^{\ln 2}(\frac{1}{2}(e^{-2x} - e^{2x})) dx = \frac{1}{M}(\frac{1}{4}(e^2-e^{-2}))=\frac{1}{4}(e^2-e^{-2})\) Hence, the mass of the thin plate is given by: \(M = \rho(1-\ln 2)\) And the centroid (center of mass) coordinates are: \((\bar{x},\bar{y}) = (\ln 2, \frac{1}{4}(e^2 - e^{-2}))\) Now we can sketch this centroid on our previously drawn region to indicate the location of the center of mass.