Problem 18 - Thomas Calculus

Question

Mass Find the mass of the solid region bounded by the parabolic surfaces \(z=16-2 x^{2}-2 y^{2}\) and \(z=2 x^{2}+2 y^{2}\) if the density of the solid is \(\delta(x, y, z)=\sqrt{x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

The mass of the solid is \( \frac{512\pi}{15} \).

Step 1: Define the Region of Integration

The given solid region is bounded by two surfaces: \( z = 16 - 2x^2 - 2y^2 \) (an upper paraboloid) and \( z = 2x^2 + 2y^2 \) (a lower paraboloid). The region of intersection of these two surfaces will form the bounds on the \( xy \)-plane. Set \( 16 - 2x^2 - 2y^2 = 2x^2 + 2y^2 \) to find the boundary of this region.Rearranging gives: \( 16 = 4x^2 + 4y^2 \), or \( x^2 + y^2 = 4 \). This describes a circle of radius 2 centered at the origin on the \( xy \)-plane.

Step 2: Set Up the Integral for Mass

The mass of the solid can be found using the triple integral of the density function over the region. In cylindrical coordinates, use \( x = r\cos(\theta) \), \( y = r\sin(\theta) \), and \( z = z \). The density function in cylindrical coordinates is \( \delta(r, \theta, z) = \sqrt{r^2\cos^2(\theta) + r^2\sin^2(\theta)} = r \).The bounds are:- \( r \) from 0 to 2 (the radius of the circle).- \( \theta \) from 0 to \( 2\pi \).- \( z \) from \( 2r^2 \) to \( 16 - 2r^2 \).Write the mass integral as:\[ M = \int_0^{2\pi} \int_0^2 \int_{2r^2}^{16-2r^2} r \cdot r \, dz \, dr \, d\theta \]

Step 3: Evaluate the Integral over z

First, evaluate the innermost integral with respect to \( z \):\[ \int_{2r^2}^{16-2r^2} r^2 \, dz = r^2 [z]_{2r^2}^{16-2r^2} = r^2 (16 - 2r^2 - 2r^2) = r^2 (16 - 4r^2) = 16r^2 - 4r^4 \]

Step 4: Evaluate the Integral over r

Next, integrate the result with respect to \( r \):\[ \int_0^2 (16r^2 - 4r^4) \, dr = \left[ \frac{16}{3}r^3 - \frac{4}{5}r^5 \right]_0^2 \]Calculate each term individually:- \( \frac{16}{3}(2)^3 = \frac{128}{3} \)- \( \frac{4}{5}(2)^5 = \frac{128}{5} \)So, the result of the integration is:\[ \frac{128}{3} - \frac{128}{5} \]Finding a common denominator:\[ \frac{128 \times 5}{15} - \frac{128 \times 3}{15} = \frac{640}{15} - \frac{384}{15} = \frac{256}{15} \]

Step 5: Evaluate the Integral over θ

Now, integrate with respect to \( \theta \):\[ \int_0^{2\pi} \frac{256}{15} \, d\theta = \frac{256}{15} [\theta]_0^{2\pi} = \frac{256}{15} \times 2\pi = \frac{512\pi}{15} \], which is the mass of the solid region.

Key Concepts

Triple Integral

A triple integral is a way to integrate over a three-dimensional region. It's like packing a solid into small pieces, determining the function's value at each piece, and summing these values. When computing the mass of a solid region, you use a triple integral to sum up the density multiplied by a differential volume across that region. In mathematical form, the triple integral for mass looks like this: \[ M = \iiint\delta(x, y, z) \, dV \]where \( \delta(x, y, z) \) is the density function and \( dV \) is the differential volume element.To solve such a problem, you first need to define the limits of integration for each dimension. In this exercise, we have a solid bounded by two surfaces, determining our limits. We need to set up the triple integral over these confines.Understanding how to set up and solve a triple integral is key. You will evaluate it in steps, beginning with the innermost integral and moving outward. The solution is achieved step-by-step, much like peeling layers of an onion.

Cylindrical Coordinates

Switching from Cartesian to cylindrical coordinates can simplify the integration process, especially when dealing with regions bounded by circular sections. In cylindrical coordinates, positions in space are determined using three values: radius \( r \), angle \( \theta \), and height \( z \). These coordinates are related to Cartesian coordinates by:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)
• \( z = z \)

The differential volume element in cylindrical coordinates is \( r \, dr \, d\theta \, dz \). This substitution simplifies the integration by making the limits consistent with the given geometrical shape.In this problem, the parabolic surfaces allow us to express \( z \) as a function of \( r \), simplifying again our integration. This method provides a more direct path to evaluating our triple integral, making complex boundaries simpler to manage geometrically.

Density Function

A density function describes how mass is distributed within a solid. For a problem like this, the density varies with position in space. Here, given by \( \delta(x, y, z) = \sqrt{x^2 + y^2} \), it indicates the density depends on the distance in the \( xy \)-plane from the z-axis.In cylindrical coordinates, the expression simplifies to \( \delta(r, \theta, z) = r \), since \( \sqrt{r^2\cos^2(\theta) + r^2\sin^2(\theta)} \) equals \( r \). This simplification aligns with how cylindrical coordinates naturally fit circular symmetry.The role of the density function in a triple integral is crucial: it weighs each volume element differently, reflecting how mass is distributed. By integrating the density function over the volume, you can calculate the total mass. Understanding density functions in such contexts allows you to solve numerous real-world engineering and physics problems.