Problem 12 - Essential Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line \(y=4\). $$ y=\frac{1}{2} x^{3}, \quad y=4, \quad x=0 $$

Step-by-Step Solution

Answer

By performing the definite integration described in the steps, we find the volume of the solid to be \(V = \frac{256\pi}{5}\) units^3.

Step 1: Identify the Bounds

We set the equation \(y = \frac{1}{2}x^3\) equal to \(y = 4\), and then solve for x to find the bounds of the integration. We get \(x = \sqrt[3]{8}\) and \(x = 0\) as the limits of our integral.

Step 2: Express the Disk Radii

We will revolve our region about \(y = 4\), which means the radius of each disk will be \(4 - y\). We then substitute \(y = \frac{1}{2}x^3\) into the equation to get \(4 - \frac{1}{2}x^3\) as the radius.

Step 3: Express the Disk Areas

Each disk will have the area \(\pi r^2\). Substitute our expression for the radius into this formula to get the area of each disk as \(\pi (4 - \frac{1}{2}x^3)^2 \).

Step 4: Find the Volume

We add up all the disk areas from \(x = 0\) to \(x = \sqrt[3]{8}\) using the formula for the volume of a solid of revolution. So, Volume \(V = \int^{sqrt[3]{8}}_0 \pi (4 - \frac{1}{2}x^3)^2 dx\). Carry out this calculus integration to get the solid's volume.

Key Concepts

Disk Method

The Disk Method is a useful technique for finding the volume of a solid of revolution. Imagine slicing the solid into thin disks perpendicular to the axis of rotation. The volume of each disk can be found and then added up, giving you the total volume of the solid.
This is done by integrating across the desired range.

• Each disk has a radius determined by the function being revolved.
• The volume of each disk is calculated using the formula for the volume of a cylinder: \( \pi r^2 h \), where \( r \) is the radius and \( h \) is the thickness of the disk, \( dx \).

This method works best when the cross-sections perpendicular to the axis of revolution are circular.

Integration Bounds

Integration bounds are the limits between which you integrate to find a total area or volume. In the context of solving for volumes of solids of revolution, these bounds represent the range where the solid exists.
To find the bounds:

• Identify the points where the revolving shape meets any bounding lines or curves, such as where \( y = \frac{1}{2}x^3 \) equals \( y = 4 \).
• In this exercise, the bounds are from \( x = 0 \) to \( x = \sqrt[3]{8} \), giving us the complete section of the curve being rotated.

These boundaries define the length over which we integrate to find the total volume.

Definite Integral

A definite integral is a way of calculating the net area under a curve. For volumes of solids of revolution, it helps sum up the infinite number of disk volumes across a defined interval.
The integral \( \int_{a}^{b} f(x) \, dx \) represents:

• Integration from lower bound \( a \) to upper bound \( b \).
• \

Cubed Function

Cubed functions involve terms raised to the third power, such as \( y = \frac{1}{2}x^3 \). They can create interesting shapes when graphed and are common in calculus exercises involving rotational solids.
When dealing with cubed functions:

• These functions can represent volumes sensitive to rotation as they create curved profiles.
• As you rotate a function like \( y = \frac{1}{2}x^3 \) around a line such as \( y = 4 \), you envision how the graph would sweep through space, forming a unique solid.

The cubed term gives the shape its distinct curves, making it exciting to explore in mathematical visualizations.