Problem 40 - Calculus of a Single Variable

Question

Comparing Integrals In Exercises 39 and \(40,\) give a geometric argument that explains why the integrals have equal values. $$ \pi \int_{0}^{2}\left[16-(2 y)^{2}\right] d y=2 \pi \int_{0}^{4} x\left(\frac{x}{2}\right) d x $$

Step-by-Step Solution

Answer

The integrals have equal values because they represent the volumes of different shapes (a half-circle and a cone) that have equivalent dimensions (base = 4 and height = 4).

Step 1: Interpret the First Integral Geometrically

Let's look at the first integral, \( \pi \int_{0}^{2}\left[16-(2 y)^{2}\right] d y \). This integral can be interpreted as the volume of a solid of revolution of the region bounded by the curve \( y = \sqrt{16 - (2x)^2} \), \( y = - \sqrt{16 - (2x)^2} \), and \( x = 2 \) about the y-axis. It is a half circle with a radius of 4, extending from 0 to 2.

Step 2: Interpret the Second Integral Geometrically

Next, let's look at the second integral, \( 2 \pi \int_{0}^{4} x\left(\frac{x}{2}\right) d x \). It represents the volume of a solid of revolution of the region bounded by the curve \( y = \frac{x^2}{2} \), \( y = 0 \), and \( x = 4 \) about the x-axis. It is a cone with base radius 4 and height 4.

Step 3: Comparing Both Integers Geometrically

Now, comparing both integrals together, the shape of the cone and the half-circle appear to be different. However, the volumes they represent are the same. This is because both the half-circle and the cone have the same base and height.

Key Concepts

Geometric Interpretation

When interpreting integrals geometrically, we delve beyond mere calculations and visualize the concepts they represent. Consider geometric shapes like circles, cones, and cylinders. These shapes often form the basis of interpreting integrals visually.
The given integrals in the exercise represent volumes of two distinct geometric figures. The first one relates to a half-circle, while the second refers to a cone. Although different, they each carve out spaces in three dimensions—a sphere with a circular base and a cone with a triangular side.
Using geometry to interpret integrals helps:

• Identify the dimensions involved: length, area, and volume.
• Understand the significance of the integrals in real-world contexts.
• Visualize mathematical computations in a tangible form.

This is crucial for grasping why integrals with different shapes may yield equal values.

Volume of Solids of Revolution

The volume of solids of revolution is a concept where a 2D shape revolves around an axis to form a 3D solid. This idea is vital to understanding how the given integrals relate geometrically.
For the first integral, the region represents a half-circle revolving around the y-axis. This creates a spherical segment. By understanding the dimensions of this shape—diameter, radius, and height—we can comprehend its volume.
On the other hand, the second integral describes a parabola revolving around the x-axis, forming a cone. The cone is characterized by its circular base and pointed top, with the radius of the base matching the height, making it a right cone.
By determining these volumes through integration, it becomes clear that:

• Both shapes are derived through continuous rotation around an axis.
• Volume calculations leverage symmetry and known geometrical formulas (e.g., for cones and spheres).

Thus, integrals can encapsulate complex volume computations in elegant expressions.

Integration

Integration is a fundamental concept in calculus, often introduced as the method of summing infinitely small areas or volumes. When applying integration to find volumes of solids of revolution or other areas, it is crucial to understand what it achieves.
In the context of the given integrals:

• The first integral employs integration to determine the total volume of the half-circle revolving to form a sphere-like shape.
• The second integral uses integration to determine the cone's volume by summing all infinitesimal discs that comprise the cone.

Integration essentially accumulates these infinitesimal pieces to provide the complete measure over a given interval.
This accumulation helps reveal that seemingly different shapes like a spherical segment and a cone can have equal volumes when determined correctly. Integration's power lies in its ability to handle these complex sums precisely and succinctly.

Definite Integrals

Definite integrals are computed over a specified interval, providing a numeric value as the result. They are an application of the integration process, utilized to find areas under curves or volumes when extended into three dimensions.
In the exercise provided:

• The first definite integral computes the volume from y=0 to y=2 of the half-circle formed by revolving around the y-axis.
• The second definite integral likewise calculates the cone's volume from x=0 to x=4 as it revolves around the x-axis.

Definite integrals are ideal for these computations because they consider all slices or discs within the specified range, which ultimately determine the total volume.
This highlights the vital role definite integrals play in translating geometric figures into calculable volumes, demonstrating that different shapes can lead to equivalent volume computations when viewed through the lens of calculus.