Problem 4 - Calculus: Early Transcendentals

Question

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. \( y = x^3 \) , \( y = 0 \) , \( x = 1 \) , \( x = 2 \)

Step-by-Step Solution

Answer

Volume is \( \frac{62\pi}{5} \).

Step 1: Setup the Problem using the Cylindrical Shells Method

To use the cylindrical shells method, consider rotating a vertical strip in the given region around the y-axis. The height of the strip is the function value, which is \( y = x^3 \), and the width of the strip is \( dx \). The distance from the y-axis to the strip (radius of the shell) is \( x \).

Step 2: Write the Formula for the Volume of the Shell

The volume \( V \) of each cylindrical shell is given by the formula: \[ V = 2 \pi \int_{a}^{b} (\text{radius}) \times (\text{height}) \times (\text{thickness}) \, dx \]For this problem:\[ V = 2 \pi \int_{1}^{2} x \cdot x^3 \, dx \]

Step 3: Simplify the Integrand

Multiply \( x \) and \( x^3 \) to get \( x^4 \). Thus, the integral becomes:\[ 2 \pi \int_{1}^{2} x^4 \, dx \]

Step 4: Integrate the Expression

Find the antiderivative of \( x^4 \), which is \( \frac{x^5}{5} \). So, the integral becomes:\[ 2 \pi \left[ \frac{x^5}{5} \right]_{1}^{2} \]

Step 5: Evaluate the Definite Integral

Calculate the antiderivative at the bounds \( x = 2 \) and \( x = 1 \):\[ \left. \frac{x^5}{5} \right|_{1}^{2} = \frac{2^5}{5} - \frac{1^5}{5} = \frac{32}{5} - \frac{1}{5} = \frac{31}{5} \]

Step 6: Find the Volume

Multiply the result from Step 5 by \( 2\pi \) to find the volume:\[ V = 2\pi \times \frac{31}{5} = \frac{62\pi}{5} \]

Key Concepts

Volume of Revolution

When we talk about the volume of revolution, we mean creating a three-dimensional shape by rotating a region or curve around an axis. Visualize this like spinning a flat surface. Everything you rotate around will form a shape in the space around the axis.
In our exercise, the region is bounded by the curves, and we spin this around the y-axis. The curve, according to the problem, is the function given by:

• \( y = x^3 \)

This means we trace out a solid, revolving it, and need to find the volume of this new shape. It is much like spinning a potter's wheel and a lump of clay forms a round object.
To calculate this volume accurately, we use a technique involving cylindrical shells. It simplifies the process by considering simpler three-dimensional pieces.

Definite Integral

A definite integral is a mathematical tool to find the total size or value between an upper and lower boundary. It helps us calculate things like the total area under a curve or, in our case, the total volume of a 3D shape.
For our given problem, we're starting and ending at specific points - here, from \( x = 1 \) to \( x = 2 \).
This range is crucial. It acts like slicing the object from edge to edge and tells us how much of the curve to include in the volume.

• When setting up a definite integral for the volume of revolution, we have: \[ V = 2\pi \int_{1}^{2} x imes (x^3) \, dx \]

This integral packs all the small cylindrical shells into one expression, pulling slices of the shape together into one whole volume.

Calculating Volumes

To calculate volumes like the one in our exercise, we must think in terms of layers. We add up the volume of each tiny piece to get the entire volume.
By using the method of cylindrical shells, each shell represents a small piece of the total volume. The shell's volume is calculated by its radius, height, and thickness.

• The radius is the distance from the axis of rotation, given by \( x \).
• The height is the value of the function, \( x^3 \).
• The thickness is the differential element, \( dx \), an infinitesimal width of these layers.

Putting them together:

• The integral \[ 2\pi \int x imes x^3 \, dx \]
• Simplifies to \[ 2 \pi \int x^4 \, dx \]

This represents the sum of all these layers from \( x=1 \) to \( x=2 \).

Integration Techniques

Integration techniques are like tools to solve different kinds of integration problems.
They're necessary for converting complex expressions into simpler ones we can evaluate or solve.
In our problem, after setting up the integral using cylindrical shells, we need to perform the integration to find the volume.

• **Multiply before integrating:** Always combine terms like \( x \) with \( x^3 \) into \( x^4 \) to simplify the problem.
• **Find the antiderivative:** Here, we use the basic power rule for integrals:The antiderivative of \( x^4 \) is \( \frac{x^5}{5} \).
• **Evaluate at boundaries using the fundamental theorem of calculus:** Make calculations at the upper and lower bounds to get the resultant volume.Plug values into \[ 2\pi \left[ \frac{x^5}{5} \right]_{1}^{2} \]
• **Simplify and calculate:** Finally, work out \[ 2\pi \times \frac{31}{5} = \frac{62\pi}{5} \]

This last step wraps up the entire process, showing how all techniques and calculations pull together to find a final result.