Problem 121 - Calculus Volume 1

Question

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the x-axis and are rotated around the y-axis. $$y=5 x^{3}, x=0, \text { and } x=1$$

Step-by-Step Solution

Answer

The volume of the solid is \( 2\pi \).

Step 1: Understand the Shell Method Formula

The shell method for finding the volume of a solid of revolution is given by the integral formula \( V = 2 \pi \int_{a}^{b} p(x) h(x) \, dx \). Here, \( p(x) \) is the distance from the axis of rotation (y-axis in this case) to the shell, and \( h(x) \) is the height of the shell.

Step 2: Identify the Functions for the Problem

For this problem, the function to be rotated is \( y = 5x^3 \) over the interval \( x = 0 \) to \( x = 1 \). The axis of rotation is the y-axis. Therefore, \( p(x) = x \), which is the distance from the y-axis to \( x \), and \( h(x) = 5x^3 \), the height of the curve above the x-axis.

Step 3: Set Up the Integral

Using the shell method, the formula becomes \( V = 2\pi \int_{0}^{1} x (5x^3) \, dx \). Simplifying inside the integral gives \( V = 2\pi \int_{0}^{1} 5x^4 \, dx \).

Step 4: Integrate the Function

To solve the integral \( \int_{0}^{1} 5x^4 \, dx \), first find the antiderivative of \( 5x^4 \), which is \( x^5 \). Therefore, the integral evaluates to \( \left[ x^5 \right]_{0}^{1} \).

Step 5: Evaluate the Integral

Plug in the limits of integration: \( \left[ 1^5 - 0^5 \right] = 1 - 0 = 1 \).

Step 6: Calculate the Final Volume

The volume is then \( V = 2\pi (1) = 2\pi \).

Key Concepts

Volume of Solids of Revolution

The concept of the volume of solids of revolution often comes into play when dealing with shapes formed by rotating a region around a specific axis. Imagine a two-dimensional area on a graph that spins around an axis, forming a three-dimensional solid. This is where we use calculus to find the volume of such solids.

There are mainly two methods to compute the volume of these revolving solids: the disk method and the shell method. In this exercise, we focus on the shell method.

• Disk Method: Involves slicing the solid in perpendicular circular disks along the axis of revolution.
• Shell Method: Involves cylinders, or "shells," sliced parallel to the axis of rotation.

The shell method is particularly useful when the region is rotated around a vertical axis (like the y-axis in this case). It is often preferred for shapes that are difficult to divide into simple disks but are 'cylindrical' in nature.

In our given problem, the region under the curve from x=0 to x=1 is rotated around the y-axis to form a solid. Using the shell method helps us comfortably deal with this vertical rotation challenge.

Integral Calculus

Integral calculus is the branch of mathematics that deals with integrals and their properties. It's one half of calculus, with the other half being differential calculus. The main goal is to find a function's antiderivative, which essentially allows for the calculation of the area under a curve. This area calculation is central to understanding shapes and curves analytically.

When we deal with volumes, like in our exercise, integral calculus comes into play powerfully:

• Definite Integrals: They provide a way to compute the exact area between a curve and an axis over a specified interval. In our case, the definite integral from x=0 to x=1 helps compute the volume within those bounds.
• Shell Method Formula: Here, the integral formula for shells, \( V = 2 \pi \int_{a}^{b} p(x) \, h(x) \, dx \), applies.

The steps involve analyzing the curve's equation, setting boundaries, and then applying the shell method formula. Mastering integral calculus improves problem-solving strategies for a broad range of real-world applications.

Antiderivatives

An antiderivative is essentially the reverse process of differentiation. When you have a function and you wish to find another function whose derivative would return you to the original, you look for its antiderivative.

In the problem we're tackling, finding the antiderivative of \( 5x^4 \) is crucial:

• The antiderivative of \( x^n \) is given by \( \frac{x^{n+1}}{n+1} \, \text{for any real number \ } n \).
• So for \( 5x^4 \, \), the antiderivative will be found as \( x^5 \, \).

Understanding and applying antiderivatives allows us to evaluate definite integrals and hence, compute volumes in calculus problems. Obtaining the antiderivative simplifies the complex integration process, letting us efficiently find the values required in real-world and theoretical calculations.