Problem 22 - Calculus and its applications

Question

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(y\) -axis. $$ y=x^{3}, x=0, x=3 $$

Step-by-Step Solution

Answer

The volume is \( \frac{486 \pi}{5} \) cubic units.

Step 1: Understand the Problem

We need to find the volume of the solid formed by rotating the region bounded by the graph of the function \( y = x^3 \), from \( x = 0 \) to \( x = 3 \), around the \( y \)-axis.

Step 2: Set Up the Integration

Since we are rotating around the \( y \)-axis, we use the method of cylindrical shells. The formula for the volume using cylindrical shells is: \[ V = 2 \pi \int_{a}^{b} x \, f(x) \, dx \] where \( f(x) = x^3 \), \( a = 0 \), and \( b = 3 \).

Step 3: Substitute and Simplify

Substitute the given function into the integral: \[ V = 2 \pi \int_{0}^{3} x \, (x^3) \, dx \]Simplify the expression inside the integral:\[ V = 2 \pi \int_{0}^{3} x^4 \, dx \]

Step 4: Integrate the Function

Calculate the integral of \( x^4 \) with respect to \( x \):\[ V = 2 \pi \, \left[ \frac{x^5}{5} \right]_{0}^{3} \]Evaluate the definite integral by substituting the limits 0 and 3.

Step 5: Evaluate the Definite Integral

First, substitute the upper limit 3:\[ \frac{3^5}{5} = \frac{243}{5} \]Then, substitute the lower limit 0, which is 0. Subtract to find:\[ V = 2 \pi \, \left( \frac{243}{5} - 0 \right) = 2 \pi \, \frac{243}{5} \]

Step 6: Final Calculation and Result

Multiply out the expression to get the final volume:\[ V = \frac{486 \pi}{5} \] Therefore, the volume of the solid is \( \frac{486 \pi}{5} \) cubic units.

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a powerful technique for finding the volume of solids of revolution. When a region in the plane is rotated around an axis to create a three-dimensional shape, this method uses cylindrical 'shells' to approximate and calculate the volume.
This method is particularly useful when the solid results from rotating a region around a vertical or horizontal line that is not the axis of the independent variable. Use it especially when rotating around the y-axis, and direct integration involving horizontal slices would be inconvenient. The shells method works well when the function is in terms of x, and the axis of rotation is parallel to the y-axis.
The formula for the volume using this method is:

• \[ V = 2 \pi \int_{a}^{b} x \, f(x) \, dx \]

Here, each shell has a height determined by the function \( f(x) \), radius \( x \) (the distance from the axis of rotation), and thickness \( dx \). The integral evaluates these shells' contributions to find the total volume of the solid.

Definite Integral

The definite integral is a fundamental concept in calculus, used to calculate the accumulation of quantities, such as areas, volumes, and other physical properties. Unlike an indefinite integral, it results in a specific numerical value rather than a general function plus a constant. When evaluating a definite integral, you consider the integral over a specific interval.
In the context of the cylindrical shells method, the definite integral provides the exact volume of a solid of revolution over a given interval \([a, b]\). The process involves setting up an integral based on the problem's requirements and evaluating it between the fixed limits of integration. In our example, we integrate from \( x=0 \) to \( x=3 \), calculating the volume created when the graph of \( y = x^3 \) rotates around the y-axis.
After finding the antiderivative of the integrand, you substitute the upper and lower limits into this expression to compute the final value of the definite integral. This subtraction (upper minus lower) completes the process, providing the volume needed in a neat numerical form, confirming the solid's volume is \( \frac{486 \pi}{5} \) cubic units.

Integration Techniques

Integration is an essential skill in calculus, and the cylindrical shells method requires careful use of these techniques. Fundamental integration involves techniques for handling polynomials, trigonometric functions, exponentials, and more complex function forms.
For the exercise, the goal is to integrate \( x^4 \). Using basic power rule integration, the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Thus, integrating \( x^4 \) gives:

• \[ \int x^4 \, dx = \frac{x^5}{5} + C \]

As it's a definite integral, the constant \( C \) cancels out. We substitute the limits of integration into \( \frac{x^5}{5} \) to find the actual volume.
Applying the integration techniques accurately is crucial for solving the problem efficiently and obtaining the correct solid volume. Always remember that interpreting the integral results in the context of the original geometric problem helps ensure the solution makes sense.