Problem 291 - Calculus Volume 1

Question

Find the generalized center of mass between \(y=a^{2}-x^{2}, \quad x=0,\) and \(y=0 .\) Then, use the Pappus theorem to find the volume of the solid generated when revolving around the \(y\) -axis.

Step-by-Step Solution

Answer

The center of mass is \( \left(\frac{3a}{8}, \frac{3}{5}\right) \) and the volume is \( \pi a^4 \).

Step 1: Understand the Problem

We need to find the center of mass of the region bounded by the curve \( y = a^2 - x^2 \), the line \( x = 0 \), and \( y = 0 \). After that, we will use Pappus's Centroid Theorem to find the volume generated when this area is revolved around the \( y \)-axis.

Step 2: Find the Limits of Integration

The intersection points will give the limits for integration. Set \( y = a^2 - x^2 = 0 \), then \( x = \pm a \). Therefore, the region of integration is from \( x = 0 \) to \( x = a \), considering symmetry, only half of the parabola is needed (from \( x = 0 \) to \( x = a \)).

Step 3: Set Up Integral for Mass

The area is given by integrating the top function minus the bottom function. Here, the function is \( y=a^2 - x^2 \). The mass \( M \) is:\[M = \int_{0}^{a} (a^2 - x^2) \, dx\].

Step 4: Calculate the Mass

Solve the integral for the mass \( M \):\[M = \int_{0}^{a} (a^2 - x^2) \, dx = \left[a^2x - \frac{x^3}{3}\right]_0^a = a^3 - \frac{a^3}{3} = \frac{2a^3}{3}\].

Step 5: Find x-Coordinate of Center of Mass

The x-coordinate of the center of mass \( \overline{x} \) is given by the formula:\[\overline{x} = \frac{1}{M} \int_{0}^{a} x(a^2 - x^2) \, dx\].

Step 6: Solve for x-Coordinate of Center of Mass

Calculate the integral for \( \overline{x} \):\[\overline{x} = \frac{3}{2a^3} \int_{0}^{a} (a^2x - x^3) \, dx = \frac{3}{2a^3} \left[ \frac{a^2x^2}{2} - \frac{x^4}{4} \right]_0^a = \frac{3}{2a^3} \left( \frac{a^4}{2} - \frac{a^4}{4} \right) = \frac{3}{2a^3} \cdot \frac{a^4}{4} = \frac{3a}{8}\].

Step 7: Find y-Coordinate of Center of Mass

The y-coordinate of the center of mass \( \overline{y} \) is invariant due to symmetry in this case as it's aligned with the y-axis. For vertical slices symmetric about the y-axis, it's \( \frac{a^2}{2} \). However, recalculating as:\[\overline{y} = \frac{1}{M} \int_{0}^{a} \frac{(a^2 - x^2)^2}{2} \, dx\].

Step 8: Calculate y-Coordinate of Center of Mass

Evaluate the definite integral for \( \overline{y} \):\[\overline{y} = \frac{3}{2a^3} \left( \frac{(a^2)x - \frac{x^3}{3}}{3} \right) = \frac{3}{5} \left[a^2x - \frac{x^3}{3}\right]_0^a = \frac{3}{5}(1) = \frac{3}{5}\].

Step 9: Apply Pappus's Centroid Theorem

Pappus's Centroid Theorem states that the volume of revolution is \( V = 2\pi R A \), where \( R \) is the x-coordinate of the center of mass and \( A \) is the area. Here \( R = \frac{3a}{8} \), and \( A = \frac{2a^3}{3} \) so:\[V = 2\pi \left( \frac{3a}{8} \right) \left( \frac{2a^3}{3} \right) = \frac{\pi a^4}{2}\].

Step 10: Final Calculation

Substitute and simplify to find the final volume using the computed center of mass coordinates:\[V = 2 \pi \overline{x} \cdot \text{Area} = 2\pi \cdot \frac{3a}{8} \cdot \frac{2a^3}{3} = \pi a^4\].

Key Concepts

Pappus's Centroid Theorem

Pappus's Centroid Theorem provides a powerful method for calculating the volume of solids of revolution. This theorem states that if a plane curve C is revolved around an external axis that does not intersect the curve, the volume of the solid formed is the product of the area of the region swept and the distance traveled by the centroid.

In simpler terms, think of the centroid as the "balance point" of a shape. When the entire region revolves around an axis, every point on this centroid circles around, forming the path needed to calculate the volume.

• First, determine the centroid (center of mass) of your region.
• Then, calculate the area of this region.
• Multiply the area by the circumference of the circle formed by the centroid to get the volume.

This method is particularly useful because it reduces a complex 3D problem to a 2D area calculation and a simple multiplication. Once you grasp the concept of centroids, applying this theorem becomes much more intuitive.

Volume of Revolution

Finding the volume of a solid formed by rotation involves using calculus techniques to sum infinitely thin disks or washers. This process is known as finding the volume of revolution and stems from revolving a 2D shape around an axis.

The method typically involves an integral setup to calculate these sums:

• Identify the axis of rotation (e.g., the x-axis or y-axis).
• Set up an integral with the limits determined by the boundaries of your region.
• Use the disk method (if rotating around the x or y-axis) or the washer method (if there's a hole) to determine the formula for the integrand.

For example, using the disk method for a function revolved about the x-axis, the volume integral is given by:\[ V = \int_{a}^{b} \pi [f(x)]^2 dx \]This method essentially sums the volumes of an infinite number of disks that make up the solid.

Calculating Integrals

Integrals allow us to calculate areas under curves, among other things. They are a fundamental concept in calculus used in computing centered on finding exact quantities, such as areas, volumes, or total mass.

To effectively work with integrals, remember the following steps:

• Identify the function you’re integrating (the integrand).
• Determine the limits of integration, which typically come from the bounds of the region being analyzed.
• Apply integration techniques such as substitution or integration by parts to solve the integral if necessary.
• Evaluate the integral at its bounds to find the desired quantity.

For instance, in the original exercise, the mass was calculated by integrating:\[ M = \int_{0}^{a} (a^2 - x^2) \, dx \]This integral measures the area under the parabolic section, which directly relates to the mass of the region.

Symmetry in Calculus

Symmetry in calculus greatly simplifies problem-solving. When a function or the region it encloses is symmetric, certain properties, such as averages or centroids, can often be deduced more easily.

Symmetry helps in:

• Halving calculations by reducing the domain of integration.
• Inferring that the centroid or center of mass will have certain coordinates purely based on symmetry.
• Fundamental properties that make evaluating integrals simpler and operations more efficient.

For instance, if a region is symmetric about the y-axis, as in our exercise, the y-coordinates of elements balanced around this axis lead directly to the center of mass being positioned on the symmetry line. This concept reduces complexity and prevents redundancy when setting up and evaluating integrals.