Question

Find the mass/weight of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\). \(R\) is the triangle with corners \((0,0),(1,0),\) and (0,1)\(;\) \(\delta(x, y)=\left(x^{2}+y^{2}+1\right) \mathrm{Ib} / \mathrm{in}^{2}\)

Short answer

The mass of the lamina is \(\frac{2}{3}\) lbs.

Step-by-step solution

Identify the Region of Integration

We are given the region \( R \) as a triangle with vertices at \((0,0), (1,0),\) and \((0,1)\). This forms a right triangle in the coordinate plane, where \(x\) and \(y\) range from 0 to 1. Specifically, \(y\) must be less than or equal to \(1-x\) within this region. We will integrate over \(R\) using limits \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\).

Set Up the Double Integral

The mass \(M\) of the lamina can be found by integrating the density function \(\delta(x, y) = x^2 + y^2 + 1\) over the region \(R\). This is expressed as: \[M = \int_{0}^{1} \int_{0}^{1-x} (x^2 + y^2 + 1) \, dy \, dx\]

Integrate with Respect to y

Start the integration by solving the inner integral, which is with respect to \(y\): \[\int_{0}^{1-x} (x^2 + y^2 + 1) \, dy. \]Split this integral into three parts: \[ \int_{0}^{1-x} x^2 \, dy + \int_{0}^{1-x} y^2 \, dy + \int_{0}^{1-x} 1 \, dy. \]Calculate each part: 1. \(x^2 \int_{0}^{1-x} \, dy = x^2[ y ]_{0}^{1-x} = x^2(1-x)\).2. \(\int_{0}^{1-x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1-x} = \frac{(1-x)^3}{3}\).3. \(\int_{0}^{1-x} 1 \, dy = [ y ]_{0}^{1-x} = 1-x\).Therefore, the result of the inner integration is:\(x^2(1-x) + \frac{(1-x)^3}{3} + (1-x)\).

Integrate with Respect to x

Substitute the result of the inner integral back into the outer integral:\[\int_{0}^{1} \left( x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \right) \, dx\]Simplify and split it into three integrals:1. \(\int_{0}^{1} x^2(1-x) \, dx \).2. \(\int_{0}^{1} \frac{(1-x)^3}{3} \, dx \).3. \(\int_{0}^{1} (1-x) \, dx \).Calculate each:1. \(\int_{0}^{1} x^2(1-x) \, dx = \int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\).2. \(\frac{1}{3}\int_{0}^{1} (1-x)^3 \, dx = \frac{1}{3} \left[ \frac{1}{4}(1-x)^4 \right]_{0}^{1} = \frac{1}{12}\).3. \(\int_{0}^{1} (1-x) \, dx = [ x - \frac{x^2}{2} ]_{0}^{1} = 1 - \frac{1}{2} = \frac{1}{2}\).Thus, adding these gives the mass:\[\frac{1}{12} + \frac{1}{12} + \frac{1}{2} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} \text{ lbs}\]

Conclusion

The mass of the lamina is the sum of the integrals from the previous step, which equates to \(\frac{2}{3}\) lbs. This result concludes the process by finding the total weight of the lamina over its specified region with the given density function.

Key concepts

Double Integral

A double integral is like an extension of a single integral into two-dimensional space, allowing us to find things like area, volume, or even mass. When working in two dimensions, we calculate a double integral to accumulate values over a specific region. In our exercise, we use a double integral to find the mass of a lamina described by a set region.

• First, double integration involves two stages of integration: one with respect to one variable and then with respect to the other.
• In our example, we start by integrating with respect to the variable \( y \), then \( x \), but the order can be switched based on ease and function complexity.
• The double integral ensures that every small area in the region contributes to the total, weighted by a given density function.

Visualizing this is like looking at a terrain map where each small patch (denoted by tiny divisions of \( x \) and \( y \)) has a specific height or density value that sums up to give us a meaningful total, such as mass.

Density Function

The density function is a critical component of finding the mass of a lamina. It represents how much mass per unit area exists at any given point, \( (x, y) \), in the defined region. In our scenario, the density function is \( \delta(x, y) = x^2 + y^2 + 1 \).

• The density function essentially tells us how 'heavy' or 'dense' each tiny part of the lamina is.
• This function needs to be integrated over the given region because the density varies for different combinations of \( x \) and \( y \).
• Functions can be constant, implying uniform density, or variable as in our case, representing more complex distributions across the region.

Recognizing the importance of the density function helps solidify understanding of why integration is necessary: we are summing up all those infinitesimal contributions to get an accurate measure of total mass.

Region of Integration

Specifying and understanding the region of integration is essential because it defines where the double integral will take its measurements. Here, the region \( R \) is a right triangle in the coordinate plane, bounded by points \((0,0), (1,0), \) and \((0,1) \).

• The region is the 'sandbox' where all the calculations happen.
• For this example, the boundaries are given by linear equations, configuring the limits of integration for each variable.
• Correctly identifying this region ensures only the intended area is considered in the calculation, preventing miscalculation.

One must carefully articulate the limits of integration, \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 - x \), to fully capture the intended triangular region and correctly apply the double integral.