Problem 77 - Calculus Volume 1

Question

For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the \(x\) -axis. \(y=x^{4}, x=0,\) and \(y=1\)

Step-by-Step Solution

Answer

The volume is \(\frac{\pi}{9}\).

Step 1: Identify the Bounded Region

The region is bounded by the curve \(y = x^4\), the line \(x = 0\), and the line \(y = 1\). This means we consider the area under the curve \(y = x^4\) from \(x = 0\) to the point where \(y = 1\). Solving for \(x\), \(x^4 = 1\) gives \(x = 1\). The region is bounded by \(x = 0\), \(x = 1\), and \(y = 0\) to \(y = 1\).

Step 2: Determine Setup for Disk Method

Since the area is rotated around the \(x\)-axis, we use the disk method. The radius of the disk is given by the distance from the \(x\)-axis to the curve \(y = x^4\), which is \(r(x) = x^4\).

Step 3: Setup the Integral

The volume \(V\) is found by integrating the area of the cross-sectional disk along the \(x\)-axis from \(x = 0\) to \(x = 1\). Hence, the integral is: \[ V = \int_0^1 \pi (x^4)^2 \, dx \] This simplifies to: \[ V = \pi \int_0^1 x^8 \, dx \]

Step 4: Evaluate the Integral

Evaluate the integral \(\int_0^1 x^8 \, dx\):1. The antiderivative of \(x^8\) is \(\frac{x^9}{9}\).2. Evaluate from 0 to 1: - \[ \left[ \frac{x^9}{9} \right]_0^1 = \frac{1^9}{9} - \frac{0^9}{9} = \frac{1}{9} \]3. Therefore, the volume is: - \[ V = \pi \cdot \frac{1}{9} = \frac{\pi}{9} \]

Step 5: Conclusion

The volume of the solid formed by rotating the area bounded by \(y = x^4\), \(x = 0\), and \(y = 1\) around the \(x\)-axis is \(\frac{\pi}{9}\).

Key Concepts

Volume of Revolution

When we talk about the volume of revolution, we're essentially referring to the volume of a 3D object formed by rotating a 2D region around a specified axis. Imagine spinning a flat shape around a line, and you'll get a 3-dimensional figure! In our exercise, this happens when the region under the curve of the equation is revolved around the x-axis, creating a solid shape.
The primary purpose of finding the volume of revolution is to determine the volume of this solid shape. This concept is particularly useful in fields such as engineering and physics, where predicting how a shape behaves is crucial.
In our specific case, we're focusing on the bounded region formed by the curve \(y = x^4\), from \(x = 0\) to \(x = 1\), and bounded above by \(y = 1\). By rotating this area around the x-axis, we construct a solid resembling a vase, but mathematically speaking, it becomes a solid of revolution.

Integrals

Integrals play a fundamental role when calculating volumes of revolution. An integral is a mathematical tool used to sum up an infinite number of tiny quantities to find areas, volumes, and more.
In the disk method, the integral helps us to add up the volume of very thin slices of the solid, which are shaped like disks. These disks are stacked along the axis of rotation, and when combined, they compose the entire volume.
For our exercise, the integral set up is: \[ V = \int_0^1 \pi (x^4)^2 \ dx \]This integral enables us to compute the volume by calculating the sum of all the small disk areas and multiplying them by the width \(dx\).
Evaluating this integral requires finding the antiderivative, which is a function whose derivative is the original function. In this case, solving:\[ \int_0^1 x^8 \ dx \] results in \(\frac{x^9}{9}\), leading to the final volume.

Solid of Revolution

A solid of revolution is formed when a 2D shape is revolved around an axis. This axis could be the x-axis, y-axis, or any other straight line. The shape of the solid reflects the boundary and nature of the 2D region before rotation. In our case, the original shape under the curve becomes a symmetric, continuous 3D object when rotated about the x-axis.
Key aspects for forming a solid of revolution include:

• Identifying the region under the curve that will be rotated.
• Determining the axis of rotation which, for our problem, is the x-axis.

The disk method involves slicing this solid into thin circular sections (disks) whose volumes are calculated and summed to give the entire volume.
By understanding the characteristics of the solid of revolution, we can apply the disk method to similar problems—estimating the volume of any object obtained by rotating a region around an axis is no longer a mystery but a mathematical procedure.