Problem 284 - Calculus Volume 1

Question

A sphere created by rotating a semicircle with radius \(a\) around the \(y\) -axis. Does your answer agree with the volume of a sphere?

Step-by-Step Solution

Answer

Yes, the volume matches the known formula \(\frac{4}{3}\pi a^3\).

Step 1: Identify the Problem

We are tasked with finding the volume of a sphere formed by rotating a semicircle around the y-axis.

Step 2: Understanding the Shape

Consider a semicircle with radius \(a\) lying in the x-y plane, with the center of its diameter on the y-axis. Rotating this semicircle around the y-axis will create a three-dimensional sphere.

Step 3: Setting Up the Integral for Volume

The volume of the sphere can be calculated using the method of disks/rings. The radius of each disk at a point "y" from the origin on the y-axis is given by \(x^2 + y^2 = a^2\), meaning \(x = \sqrt{a^2 - y^2}\).

Step 4: Integrate to Find the Volume

The volume \(V\) of the sphere by summing up the disks is given by the integral:\[ V = 2\pi \int_{-a}^{a} x \, dy = 2\pi \int_{-a}^{a} \sqrt{a^2 - y^2} \, dy \] Using symmetry, this simplifies to \[ 4\pi \int_{0}^{a} \sqrt{a^2 - y^2} \, dy \].

Step 5: Evaluate the Integral

To solve the integral, use the substitution \(y = a \sin \theta\), then \(dy = a \cos \theta \, d\theta\). When \(y = 0\), \(\theta = 0\), and when \(y = a\), \(\theta = \frac{\pi}{2}\). The integral becomes:\[ 4\pi \int_{0}^{\frac{\pi}{2}} a^2 \cos^2 \theta \, d\theta \]Solving gives the volume as \(\frac{4}{3}\pi a^3\).

Step 6: Match with Known Sphere Volume Formula

The formula \(\frac{4}{3}\pi a^3\) is indeed the formula for the volume of a sphere with radius \(a\), confirming the solution.

Key Concepts

Calculus

Calculus is a fundamental mathematical tool that helps us solve complex problems involving change and motion. In the context of finding the volume of a sphere through integration, calculus allows us to handle the intricate task of summing up infinitely small elements to find the total volume.

In this problem, calculus comes into play when setting up and evaluating the integral that calculates the volume. By utilizing techniques like substitution and integration, we can handle the infinitely many thin slices (disks) that make up the sphere when it is formed by rotating a semicircle.

• This use of calculus is especially apparent in Step 5, where a trigonometric substitution streamlines solving the integral.

These ideas form the core of why calculus is indispensable in not only geometry but also broader mathematical applications. With calculus, the transition from a plane shape to a solid becomes more manageable and systematic.

Rotational Solids

Rotational solids arise from rotating a 2D shape around an axis to generate a 3D object. In our exercise, a semicircle is rotated around the y-axis, producing a sphere. This concept links closely to the idea of symmetry and the balanced distribution of matter in geometric spaces.

To visualize this, imagine a flat semicircle in the x-y plane. When rotated around the y-axis, each point of the semicircle traces out a path in three-dimensional space, forming a continuous surface—a sphere.

• This rotation leverages the natural symmetry of circles, which ensures that the 3D sphere is uniform and perfectly round.

The beauty of this idea is seen when applying calculus' integral techniques to compute the total volume, verifying that this practical method matches theoretical expectations.

Integration Techniques

Integration techniques are essential for solving volume problems where simple geometric formulas are inadequate. In the problem of determining the volume of a sphere, we use specific integration techniques to sum the volumes of infinitely small disks that make up the sphere.

For this, the disk method is employed, involving:

• Setting up an integral based on a formula that represents the volume of thin disks with changing radii.
• The substitution method is used for handling complex integrals, seen when \(y = a \sin \theta\) was utilized to simplify the expression.
• Converting limits of integration from y-space to θ-space, which makes the process both efficient and comprehensible.

These techniques are pivotal in extending simple geometric principles to solve more complex problems, thus demonstrating the power and flexibility of integration in calculus.