Problem 7 - Thomas Calculus

Question

The solids in Exercises \(1-12\) all have constant density \(\delta=1\) a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid \(z=x^{2}+y^{2}\) and above by the plane \(z=4\) . b. Find the plane \(z=c\) that divides the solid into two parts of equal volume. This plane does not pass through the center of mass.

Step-by-Step Solution

Answer

Center of mass is at (0, 0, 8/5); plane z = 2\sqrt{2} divides volumes equally.

Step 1: Set Up the Problem

The solid is bounded by the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 4 \). We need to find the center of mass and the plane \( z = c \) that divides the solid into two equal volumes. The density \( \delta = 1 \).

Step 2: Determine Volume

The volume \( V \) of the solid is calculated by integrating over the region between the paraboloid and the plane. The volume integral is given by:\[V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} r \, dz \, dr \, d\theta \]Evaluating this integral will give the total volume of the solid.

Step 3: Calculate Volume Integral

Calculate the integral for volume:\[V = \int_{0}^{2\pi} \int_{0}^{2} [z]_{r^2}^{4} \, r \, dr \, d\theta \]\[= \int_{0}^{2\pi} \int_{0}^{2} (4-r^2)r \, dr \, d\theta\]\[= \int_{0}^{2\pi} \left[ 2r^2 - \frac{r^4}{4} \right]_{0}^{2} \, d\theta\]\[= \int_{0}^{2\pi} \left(8 - 4\right) \, d\theta \]\[= \int_{0}^{2\pi} 4 \, d\theta = 8\pi\]

Step 4: Center of Mass Calculation

The center of mass coordinates are: \(\bar{x} = \frac{1}{V}\int x \, dV, \ \bar{y} = \frac{1}{V}\int y \, dV, \ \bar{z} = \frac{1}{V}\int z \, dV\)Since the solid is symmetric about the \( z \)-axis, \( \bar{x} = \bar{y} = 0 \). Calculate \( \bar{z} \):\[\bar{z} = \frac{1}{V}\int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} z r \, dz \, dr \, d\theta = \frac{1}{8\pi} \int_{0}^{2\pi} \int_{0}^{2} \left[ \frac{z^2}{2} \right]_{r^2}^{4} r \, dr \, d\theta\]\[= \frac{1}{8\pi} \int_{0}^{2\pi} \int_{0}^{2} \left( \frac{16}{2} - \frac{(r^2)^2}{2} \right) r \, dr \, d\theta\]\[= \frac{1}{8\pi} \int_{0}^{2\pi} [8r - \frac{r^5}{10}]_{0}^{2} \, d\theta\]\[= \frac{1}{8\pi} \int_{0}^{2\pi} (16 - \frac{32}{10}) \, d\theta = \frac{1}{8\pi} \int_{0}^{2\pi} \frac{64}{5} \, d\theta = \frac{64\pi}{40\pi} = \frac{8}{5}\]

Step 5: Plane Dividing Solid

To find the plane \( z = c \) that divides the solid into two equal volumes, we solve for \( c \) such that the volume of the solid below \( z = c \) is half of the total volume (\(8\pi/2 = 4\pi\)).\[\int_{0}^{2\pi} \int_{0}^{\sqrt{c}} \int_{r^2}^{c} r \, dz \, dr \, d\theta = 4\pi\]\[\int_{0}^{2\pi} \int_{0}^{\sqrt{c}} (c - r^2) r \, dr \, d\theta = 4\pi\]\[= \int_{0}^{2\pi} \left[ \frac{cr^2}{2} - \frac{r^4}{4} \right]_{0}^{\sqrt{c}} \, d\theta = \int_{0}^{2\pi} \left( \frac{c^2}{2} - \frac{c^2}{4} \right) \, d\theta\]\[= \int_{0}^{2\pi} \frac{c^2}{4} \, d\theta = 4\pi \Rightarrow \frac{c^2}{4} \cdot 2\pi = 4\pi \Rightarrow c^2 = 8 \Rightarrow c = 2\sqrt{2}\]

Step 6: Conclusion

The center of mass of the solid is \( (0, 0, \frac{8}{5}) \). The plane \( z = 2\sqrt{2} \) divides the solid into two parts of equal volume.

Key Concepts

Center of Mass

Imagine trying to balance an object on your fingertip. The point where it balances perfectly is known as the center of mass. In calculus, finding the center of mass involves calculating where this point exists for a 3D object. For a solid with uniform density, the process simplifies considerably because the density is the same throughout the object.
For a solid symmetric about the z-axis, the x and y-coordinates of the center of mass are zero, meaning the object balances in the center horizontally. The z-coordinate is found using integration:

• Calculate the total volume of the object.
• Use integrals to find the weighted average height of all the infinitesimally small mass elements.

This calculation provides the z-coordinate of the center of mass, telling us how high above the base it is for a symmetrical solid.

Volume of Solids

Calculating the volume of a solid is crucial in many areas of science and engineering. In this example, we find the volume of a solid bounded by a paraboloid and a plane—a shape sort of like an ice cream cone with a flat top.
To find this volume, we use integration, which allows us to add up tiny, infinite slices of the solid. We set up a triple integral:

• The innermost integral adds up heights (from the paraboloid surface up to the flat plane).
• The middle integral accumulates these slices across radii outwards from the center.
• The outer integral rotates the whole collection around the central axis.

Solving these integrals gives us the total volume, which is essential to understanding the extent of a solid.

Triple Integrals

Triple integrals are an extension of double integrals which are used to compute volumes. They integrate across three overlapping dimensions, calculating sums of small volume elements over a region in space.
In this problem, we use cylindrical coordinates, which are particularly useful for solids with circular symmetry. Cylindrical coordinates include radius (\( r \)), angle (\( \theta \)), and height (\( z \)).

• Convert the problem into cylindrical coordinates if the solid has rotational symmetry around an axis.
• Integrate step by step in each dimension, typically starting with the smallest (innermost) variable.
• Solve the integrals methodically, ensuring you keep track of the limits at each stage.

Using triple integrals allows us to carefully account for volume in complex shapes and can be expanded to concepts beyond just volume, such as total mass or charge density.

Symmetry in Solids

Symmetry can significantly simplify the process of analyzing a solid, especially when dealing with mathematical computations like finding centers of mass or dividing volumes. In a symmetrical solid, certain properties repeat or mirror in a consistent way.
For example, if a solid is symmetric around an axis, such as the z-axis in our problem, calculations along the perpendicular axes can often be simplified. This holds because symmetrical distributions mean that any influences in one direction are balanced by symmetric influences in the opposite.
Key practices when dealing with symmetrical solids include:

• Identify axes or planes of symmetry to reduce computation efforts.
• Recognize that symmetry can lead to many values (like center of mass in certain directions) becoming zero.
• Use the symmetry to simplify boundary conditions and set up integrals accordingly.

Engaging symmetry efficiently can turn complex problems into more manageable tasks, saving time and reducing potential for errors.