Problem 12 - APEX Calculus

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 rectangle with corners (1,-3),(1,2),(7,2) and (7,-3)\(; \delta(x, y)=\left(x+y^{2}\right) \mathrm{gm} / \mathrm{cm}^{2}\)

Step-by-Step Solution

Answer

The mass of the lamina is 190 gm.

Step 1: Identify the Region of Integration

The lamina's region of interest, \(R\), is a rectangle described by the corners (1,-3), (1,2), (7,2), and (7,-3). Hence, we can use the bounds for integration as \(x = 1\) to \(x = 7\) and \(y = -3\) to \(y = 2\).

Step 2: Set Up the Double Integral

The mass of the lamina is calculated by integrating the density function over the region \(R\). Thus, the mass \(M\) is given by the integral: \[ M = \int_{1}^{7} \int_{-3}^{2} \delta(x, y) \, dy \, dx = \int_{1}^{7} \int_{-3}^{2} (x + y^2) \, dy \, dx \].

Step 3: Perform Inner Integration with Respect to y

First, integrate with respect to \(y\): \[ \int_{-3}^{2} (x + y^2) \, dy = \int_{-3}^{2} x \, dy + \int_{-3}^{2} y^2 \, dy \]. Calculate each part: - \( \int_{-3}^{2} x \, dy = x[y]_{-3}^{2} = x(2 - (-3)) = 5x \).- \( \int_{-3}^{2} y^2 \, dy = \left[\frac{y^3}{3}\right]_{-3}^{2} = \frac{2^3}{3} - \frac{(-3)^3}{3} = \frac{8}{3} + 9 = \frac{35}{3} \).Thus, \( \int_{-3}^{2} (x + y^2) \, dy = 5x + \frac{35}{3} \).

Step 4: Perform Outer Integration with Respect to x

Now integrate the result from the inner integral with respect to \(x\):\[ \int_{1}^{7} \left( 5x + \frac{35}{3} \right) \, dx = \int_{1}^{7} 5x \, dx + \int_{1}^{7} \frac{35}{3} \, dx \].Calculate each part:- \( \int_{1}^{7} 5x \, dx = 5\left[\frac{x^2}{2}\right]_{1}^{7} = 5\left(\frac{49}{2} - \frac{1}{2}\right) = 5 \cdot \frac{48}{2} = 5 \times 24 = 120 \).- \( \int_{1}^{7} \frac{35}{3} \, dx = \frac{35}{3}[x]_{1}^{7} = \frac{35}{3}(7 - 1) = \frac{35}{3} \times 6 = 70 \).Finally:\[ M = 120 + 70 = 190 \].

Step 5: Interpret the Result

The total mass of the lamina is \(190 \, \textrm{gm}\), given the density function \(\delta(x, y) = x + y^2\) and the region defined.

Key Concepts

Density Function

A density function is a mathematical tool used to define a variable distribution over a given region. In the context of a lamina, it tells how mass is distributed over the area. For the exercise, the density function is given as \( \delta(x, y) = x + y^2 \). This particular density function indicates that the mass at any point \((x, y)\) increases linearly with \(x\) and quadratically with \(y\). This means that as you move away from the origin, the mass will increase either due to an increase in \(x\) or \(y\) or both.

Such density functions are not uniform unless simplified to a constant. Hence, integration over the specific region is required to find the total mass. It's crucial to clearly understand how changes in the variables \(x\) and \(y\) impact the overall density at each point of the lamina.

Region of Integration

The region of integration specifies the area over which we will integrate our density function. It is described by the boundaries in the coordinate plane. In this problem, the region \(R\) is a rectangle with vertices at \((1, -3), (1, 2), (7, 2), \text{and} (7, -3)\).

This region is crucial because it defines where the mass is located and hence where our density function applies. Setting proper boundaries is vital, as they dictate the limits of our integration process. The x-axis limits are from \(1\) to \(7\), and the y-axis limits are from \(-3\) to \(2\). These boundaries form a rectangle, allowing us to integrate properly over its area.

Inner Integration

Inner integration refers to the first integration performed in a double integral, which is usually taken with respect to one variable while treating the other variable as constant. In our example, you perform inner integration with respect to \(y\), over the interval \([-3, 2]\).

To break it down, you divide the function \(x + y^2\) into two separate integrals: \(\int_{-3}^{2} x \, dy\) and \(\int_{-3}^{2} y^2 \, dy\). Individual computations result in \(5x\) for the first part and \(\frac{35}{3}\) for the second part. When you sum these results, you obtain the expression \(5x + \frac{35}{3}\) that will be further integrated with respect to \(x\).

Understanding this step is crucial. It simplifies handling multidimensional functions, making the problem-solving more approachable.

Outer Integration

Outer integration is the subsequent integration where you handle the variable that was kept constant during the inner integration. Here, you perform it with respect to \(x\), adding detail to the previous result from the inner integration.

The integral \(\int_{1}^{7} \left( 5x + \frac{35}{3} \right) \, dx\) involves computing two separate integrals: \(\int_{1}^{7} 5x \, dx\) and \(\int_{1}^{7} \frac{35}{3} \, dx\). Evaluating them gives \(120\) and \(70\) respectively, and their sum provides the final mass \(M\) of the lamina as \(190\, \text{gm}\).
By the end of the outer integration, the entire mass across the region of integration has been accounted for using the density function across both variables. This is the last step in computing the total quantity you're looking for and highlights how double integrals accumulate value over two dimensions.