Question

Find the mass \(m\) and center of mass \((\bar{x}, \bar{y})\) of the lamina bounded by the given curves and with the indicated density. \(y=e^{-x}, y=0, x=0, x=1 ; \delta(x, y)=y^{2}\)

Short answer

Mass \(m = \frac{1}{9}(1 - e^{-3})\), center of mass \((\bar{x}, \bar{y}) = \left(\frac{9(4 - e^{-3})}{1 - e^{-3}}, \frac{9(1 - e^{-4})}{1 - e^{-3}}\right)\).

Step-by-step solution

Identify the region and density function

The region bounded by the curves is defined by the equations: \(y = e^{-x}\), \(y = 0\), \(x = 0\), and \(x = 1\). The density function is given as \(\delta(x, y) = y^2\). The region is a bounded area under the curve \(y = e^{-x}\) from \(x = 0\) to \(x = 1\).

Set up the integral for mass

The mass \(m\) of the lamina is given by the integral of the density over the region:\[m = \int_{x=0}^{1} \int_{y=0}^{e^{-x}} y^2 \, dy \, dx\]

Compute the inner integral for mass

The inner integral computes the contribution to the mass from a vertical strip at a fixed \(x\):\[\int_{y=0}^{e^{-x}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{e^{-x}} = \frac{(e^{-x})^3}{3}\]

Compute the outer integral for mass

Substitute the result from Step 3 into and compute the outer integral:\[m = \int_{x=0}^{1} \frac{e^{-3x}}{3} \, dx = \frac{1}{3} \left[ -\frac{1}{3} e^{-3x} \right]_{0}^{1} = \frac{1}{9} (1 - e^{-3})\]

Set up the integral for the x-coordinate of the center of mass

The x-coordinate of the center of mass \(\bar{x}\) is given by:\[\bar{x} = \frac{1}{m} \int_{x=0}^{1} \int_{y=0}^{e^{-x}} x y^2 \, dy \, dx\]

Compute the inner integral for \(\bar{x}\)

Evaluate the inner integral for \(\bar{x}\):\[\int_{y=0}^{e^{-x}} x y^2 \, dy = x \left[ \frac{y^3}{3} \right]_{0}^{e^{-x}} = x \frac{(e^{-x})^3}{3}\]

Compute the outer integral for \(\bar{x}\)

Substitute and evaluate the outer integral for \(\bar{x}\):\[\bar{x} = \frac{1}{m} \int_{x=0}^{1} \frac{x e^{-3x}}{3} \, dx\]Using integration by parts, solve:\[\int x e^{-3x} \, dx = -\frac{e^{-3x}}{9} (3x + 1)\]Evaluating between 0 and 1 gives:\[\int_{0}^{1} x e^{-3x} \, dx = -\frac{1}{9} (4 - e^{-3})\]Thus,\[\bar{x} = \frac{1}{m} \cdot \frac{1}{9} (4 - e^{-3})\]

Set up the integral for the y-coordinate of the center of mass

The y-coordinate is given by:\[\bar{y} = \frac{1}{m} \int_{x=0}^{1} \int_{y=0}^{e^{-x}} y^3 \, dy \, dx\]

Compute the inner integral for \(\bar{y}\)

Evaluate the inner integral for \(\bar{y}\):\[\int_{y=0}^{e^{-x}} y^3 \, dy = \left[ \frac{y^4}{4} \right]_{0}^{e^{-x}} = \frac{(e^{-x})^4}{4}\]

Compute the outer integral for \(\bar{y}\)

Substitute the inner integral and evaluate the outer integral for \(\bar{y}\):\[\bar{y} = \frac{1}{m} \int_{x=0}^{1} \frac{e^{-4x}}{4} \, dx = \frac{1}{m} \cdot \frac{1}{4} \left[ -\frac{1}{4} e^{-4x} \right]_{0}^{1} = \frac{1}{m} \cdot \frac{1}{16} (1 - e^{-4})\]

Calculate \((\bar{x}, \bar{y})\) and conclude

After calculating \(m\) from Step 4 and using it in Steps 7 and 10, we find:\[\bar{x} = \frac{(4 - e^{-3})}{(1 - e^{-3})/9} = \frac{9(4 - e^{-3})}{1 - e^{-3}}\]\[\bar{y} = \frac{(1 - e^{-4})}{(1 - e^{-3})/9} = \frac{9(1 - e^{-4})}{1 - e^{-3}}\]The mass \(m\) and the center of mass are thus determined.

Key concepts

Integral Calculus

Integral calculus is a fundamental aspect of mathematics that deals with the process of integrating functions. In simpler terms, it involves finding the accumulation, or total amount, of a quantity over a particular interval. This method is essential when we want to calculate areas under curves, like the mass of objects with varying densities, among other applications.
In our exercise, we integrated the density function over a specified region to find the mass of the lamina. The process involves setting up and calculating both vertical (inner integrals) and horizontal (outer integrals). This method allows us to accumulate small quantities across a defined region to find the overall mass, or other measures, of our shape.
The beauty of integral calculus lies in its ability to handle complex, continuous changes and find meaningful results. Without it, finding such integral parts of our universe would be quite challenging.

Lamina

A lamina is a flat, two-dimensional object with mass and a specific density function over its surface. It's similar to a very thin plate or sheet. The lamina in our problem is defined by boundaries from the functions given, forming a specific shape.
Understanding the shape of the lamina is crucial because it defines the region over which we perform our integrations. By carefully examining and plotting the boundaries - in this case, the curves where the lamina is bounded - we can better grasp how the lamina is shaped. This, in turn, helps in setting up integrals accurately for any calculations.
Visualizing lamina helps to comprehend the integration segments, ensuring no part of the shape is missed or miscalculated in the integrals.

Density Function

The density function is a mathematical formula that specifies how density varies across a region or object. In physics, the density function helps determine how mass is distributed, affecting how we calculate overall mass and the center of mass.
In the current problem, the density is given by the function \(\delta(x, y) = y^2\). This function indicates how density changes depending on the position \(y\) on the lamina. Understanding the density function allows us to incorporate these variations into our integral calculus to find accurate mass when doing computations.
By integrating this function over a designated region, we can calculate how much mass is within that surface and further move on to determining its center of mass. Keeping track of how density changes is key to precise and realistic mass computations.

Mass Calculation

Mass calculation using integral calculus revolves around integrating the density function over the given region. This method involves setting up integrals to cover the entire lamina shape made by the boundary curves.
In our example, we calculated the mass by integrating the density function \(y^2\) over the prescribed region. We first tackled the inner integral, which evaluated mass contributions from vertical strips; then moved outward to evaluate the sum from zero to one horizontally.
Our calculation yielded a mass \(m = \frac{1}{9} (1 - e^{-3})\). Getting the correct mass ensures we can accurately proceed with center of mass computations. It's a balance of understanding shape, density function, and executing integration accurately.