Problem 17 - Calculus and its applications

Question

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(x\) -axis. $$ y=\sqrt{4-x^{2}}, x=-2, x=2 $$

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{32\pi}{3}\).

Step 1: Understand the Problem

We need to find the volume of the solid formed by rotating the region bounded by the curve \(y=\sqrt{4-x^{2}}\) and the lines \(x=-2\) and \(x=2\) about the \(x\)-axis.

Step 2: Identify the Region of Revolution

The function \(y=\sqrt{4-x^{2}}\) represents the upper half of the circle with center at the origin and radius 2. The lines \(x=-2\) and \(x=2\) mark the left and right boundaries, respectively, of this semicircular region.

Step 3: Set Up the Integral for Volume

We'll use the disk method. The radius of each disk is given by the function \(y=\sqrt{4-x^{2}}\), and the breadth of an infinitesimally small disk is \(dx\). The volume \(V\) of the solid is given by the integral formula: \[ V = \int_{-2}^{2} \pi (\sqrt{4-x^{2}})^2 \, dx \] which simplifies to: \[ V = \int_{-2}^{2} \pi (4-x^{2}) \, dx \].

Step 4: Evaluate the Integral

First evaluate the integral: \[ \int (4-x^2) \, dx = 4x - \frac{x^3}{3} + C \].Now apply the definite integral from \(-2\) to \(2\): \[ V = \pi \left[ (4x - \frac{x^3}{3}) \right]_{-2}^{2} \].

Step 5: Solve the Definite Integral

Substitute the limits into the antiderivative: \[ (4 \cdot 2 - \frac{2^3}{3}) - (4 \cdot (-2) - \frac{(-2)^3}{3}) \].Calculate the values: \(\left[8 - \frac{8}{3} \right] - \left[-8 + \frac{8}{3}\right] = \left[ \frac{24}{3} - \frac{8}{3}\right] - \left[-\frac{24}{3} + \frac{8}{3}\right] = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \).

Step 6: Calculate the Final Volume

Multiply the result by \(\pi\): \[ V = \pi \cdot \frac{32}{3} = \frac{32\pi}{3} \].

Key Concepts

Disk Method

The Disk Method is a useful technique in calculus for finding volumes of solids of revolution. Imagine slicing a solid into thin disks perpendicular to an axis of rotation. Each slice resembles a disk (or a washer, if there's a hole). In our case, the region bounded by the curve rotates around the \(x\)-axis, creating these disks.To set up the Disk Method:

• Identify the function describing the radius of each disk. Here, it's \(y = \sqrt{4-x^2}\).
• The thickness of each disk is a tiny change in \(x\) (denoted as \(dx\)).
• The volume of one disk is given by \(\pi r^2\cdot dx\), where \(r\) is the radius of the disk.

To find the volume of the entire solid:

• Integrate the volume of each disk from the starting point to the endpoint of the region along the \(x\)-axis.

In this exercise, you integrate from \(-2\) to \(2\) to capture the entire semicircle, resulting in a more straightforward solution.

Definite Integral

Definite integrals are a way to calculate the accumulation of quantities, like area or volume, between two bounds. In this context, they help find the total volume formed by revolving a region around an axis.### Setting Up the IntegralFor volume using disks, the setup involves:

• Recognizing the geometric region to revolve. Here, it involves the curve between \(x = -2\) and \(x = 2\).
• The formula becomes: \[ V = \int_{a}^{b} \pi [f(x)]^2 \, dx \]
• For our function \(f(x) = \sqrt{4-x^2}\), plug it into the integral: \[ V = \int_{-2}^{2} \pi (4-x^2) \, dx \]

### Evaluating the IntegralBy performing the integral calculus:

• Find the anti-derivative of the integrand, which simplifies to \(4x - \frac{x^3}{3}\).
• Plug in upper and lower bounds into the anti-derivative to calculate the definite integral.

The result is a precise calculation of volume, showing how integrals translate geometric boundaries into concrete measures.

Circular Region

A circular region is a two-dimensional area shaped like a circle or part of a circle. In problems involving volumes of revolution, identifying this shape helps define boundaries for computation.To define our circular region:

• Understand the curve \(y = \sqrt{4-x^2}\) outlines the top half of a circle with a radius of 2, centered at the origin.
• The range from \(x = -2\) to \(x = 2\) completes this semicircle.

When this region revolves around the \(x\)-axis, it forms a three-dimensional shape, resembling a solid sphere. When tackling such problems:

• Illustrate the curve and bounds to visualize the circular region effectively.
• Recognize how these parameters contribute to defining the disk dimensions in the Disk Method.

Understanding the circular region is crucial, as it directly impacts how you compute integrals for solids of revolution.