Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

An iterated integral is given either in rectangular or polar coordinates. The double integral gives the mass of some lamina \(R\). Sketch the lamina \(R\) and determine the density \(\delta .\) Then find the mass and center of mass. \(\int_{-3}^{3} \int_{0}^{9-x^{2}} k\left(x^{2}+y^{2}\right) d y d x\)

Step-by-Step Solution

Answer

Region R is a semicircle, mass is found by integrating the given expression, center of mass involves computing similar integrals.

Step 1: Identify the Region R

The iterated integral is given as \( \int_{-3}^{3} \int_{0}^{9-x^{2}} k(x^{2}+y^{2}) \ dy \ dx \). From the given limits of integration, we realize that \( x \) ranges from \(-3\) to \(3\) and for each \( x \), \( y \) ranges from \(0\) to \(9-x^{2}\). The function \( y = 9 - x^{2} \) denotes the upper boundary, which is a downward-facing parabola intersecting the x-axis at \( x = -3 \) and \( x = 3 \). The region \( R \) lies below this parabola and above the x-axis.

Step 2: Sketch Region R

To sketch the region \( R \), draw the parabola \( y = 9 - x^2 \) in a coordinate plane. It opens downwards and has its vertex at \( (0, 9) \). The parabola intersects the x-axis at \( x = \pm 3 \), hence this is the domain along the x-axis. The area of interest is above the x-axis as y ranges from \(0\) up to the parabola.

Step 3: Determine the Density Function

The density function is given by \( \delta(x, y) = k(x^2 + y^2) \). As \( k \) is a constant, it represents a proportionality constant for the density function based on the radial distance squared from the origin \( x^2 + y^2 \).

Step 4: Find the Mass

The mass \( M \) of the lamina can be found by evaluating the given double integral. Start by integrating with respect to \( y \):\[ \int_{-3}^{3} \int_{0}^{9-x^{2}} k(x^2 + y^2) \, dy \, dx. \]Integrate the inner integral:\[ \int_{0}^{9-x^2} (k(x^2+y^2)) \, dy = k \left[ x^2y + \frac{y^3}{3} \right]_0^{9-x^2} = k \left[ x^2(9-x^2) + \frac{(9-x^2)^3}{3} \right]. \]This needs to be evaluated from \( x=-3 \) to \( x=3 \).

Step 5: Evaluate the Integral for Mass

Substitute \( y = 9-x^2 \) into the expression for mass obtained from the earlier step:\[ M = \int_{-3}^{3} k \left[ x^2(9-x^2) + \frac{(9-x^2)^3}{3} \right] \, dx. \]Calculate this by simplifying and integrating term-by-term:- For \( x^2(9-x^2) = 9x^2 - x^4 \), integrate to get \( \frac{9x^3}{3} - \frac{x^5}{5} \).- The \( \frac{(9-x^2)^3}{3} \) term is more complex, but typically involves substitution methods or numeric approximation depending on specific boundaries.Integrate over \( x = -3 \) to \( x = 3 \).

Step 6: Calculate the Center of Mass

The center of mass \( (\overline{x}, \overline{y}) \) is found using:\[ \overline{x} = \frac{1}{M} \int_{-3}^{3} \int_{0}^{9-x^2} x \cdot k(x^2+y^2) \, dy \, dx \]\[ \overline{y} = \frac{1}{M} \int_{-3}^{3} \int_{0}^{9-x^2} y \cdot k(x^2+y^2) \, dy \, dx. \]Evaluate these integrals using similar steps to those used for the mass to find \( \overline{x} \) and \( \overline{y} \).

Key Concepts

mass of lamina

The mass of a lamina is a crucial concept in calculus that is computed using a double integral. A lamina is a two-dimensional object, like a thin plate, with a certain distribution of mass across its surface. The given exercise involves evaluating the mass of such a lamina, represented by the region \( R \).
The mass \( M \) of the lamina is calculated by integrating the density function over the entire region \( R \). In the context of rectangular coordinates, this is expressed as the double integral:

• \[ M = \int \int_{R} \delta(x, y) \, dA \]

where \( \delta(x, y) \) is the density function and \( dA \) signifies the differential area element, typically \( dy \, dx \) or \( dx \, dy \) depending on the limits of integration.In the provided problem, the integral is defined with limits suggesting that the lamina extends horizontally from \( x = -3 \) to \( x = 3 \) and vertically up to the parabola \( y = 9 - x^2 \). Evaluating this integral provides the total mass by summing the contributions of density at every point in this region.

density function

Density functions describe how mass is distributed within a lamina, which is vital for calculating both mass and the center of mass. In most problems, density might vary across the lamina based on its position.The density function given in the exercise is \( \delta(x, y) = k(x^2 + y^2) \). Here, \( k \) is a constant, representing a simple case where density increases with the square of the radial distance from the origin. This is common for many materials whose density distribution depends on how far they are from a certain point.
Understanding the density function involves:

• Identifying how it changes with position in the lamina.
• Recognizing that it influences both mass calculations and the position of the center of mass.

In this scenario, the density function is quadratic in nature, suggesting that the density is higher at positions further from the origin, encompassing a symmetrical spread around the origin.

center of mass

The center of mass is the point where the entire mass of an object can theoretically be concentrated. It's crucial in physics and engineering, as it helps in predicting motion and stability of structures.For a two-dimensional lamina, the coordinates \( (\overline{x}, \overline{y}) \) of the center of mass are determined by the integrals:

• \[ \overline{x} = \frac{1}{M} \int \int_{R} x \cdot \delta(x,y) \, dA \]
• \[ \overline{y} = \frac{1}{M} \int \int_{R} y \cdot \delta(x,y) \, dA \]

These expressions show that the place where mass is distributed affects where the center of mass will be. Here, the integrals involve a weighted average of the coordinates, where the 'weight' is the density function.
For the given region \( R \), the center of mass needs to consider the density function \( k(x^2 + y^2) \) across the parabola-defined area. Therefore, the coordinates \( (\overline{x}, \overline{y}) \) are found by resolving these integrals using similar steps as for the mass, but now including terms \( x \cdot \delta(x,y) \) and \( y \cdot \delta(x,y) \).

rectangular coordinates

Rectangular coordinates, also known as Cartesian coordinates, are likely the most familiar system for plotting points and regions on a plane. They are defined by two perpendicular axes (x and y) that allow any point in the plane to be represented by a pair of numbers.In the current exercise, rectangular coordinates help in defining the limits of integration for the mass and center of mass calculations. The iterated integral provided suggests limits that transform the problem into finding the volume under a surface, thereby giving mass.For the integral

• \( \int_{-3}^{3} \int_{0}^{9-x^{2}} k(x^2 + y^2) \, dy \, dx \)

the range of the outer integral \( x = -3 \) to \( x = 3 \) and the inner integral \( y = 0 \) to \( y = 9-x^2 \) are set using rectangular coordinates, making it clear that the region is bound above by the curve \( y = 9 - x^2 \) and lies symmetric around the y-axis. Using rectangular coordinates simplifies the definition of such regions and aids in precise integration across defined bounds.