Problem 441 - Calculus Volume 1

Question

For the following exercises, use the requested method to determine the volume of the solid. \(x=y^{2}\) and \(x=3 y\) rotated around the \(y\) -axis using the washer method

Step-by-Step Solution

Answer

The volume is \(\frac{162\pi}{5}\).

Step 1: Identify the region of rotation

First, we need to identify the region bounded by the curves. The curves given are \(x = y^2\) and \(x = 3y\). To find their points of intersection, set \(y^2 = 3y\). Solving this gives \(y(y - 3) = 0\), hence \(y = 0\) or \(y = 3\). This means our region of rotation is between \(y = 0\) and \(y = 3\).

Step 2: Set up the integral for the washer method

In the washer method, the volume is calculated using concentric circular slices perpendicular to the axis of rotation. When rotating around the \(y\)-axis, we use the formula: \[V = \pi \int_a^b (\text{outer radius}^2 - \text{inner radius}^2) \, dy\]Here, \(a = 0\) and \(b = 3\). The outer radius is \(x = 3y\) and the inner radius is \(x = y^2\). Therefore, the integral becomes:\[V = \pi \int_0^3 ((3y)^2 - (y^2)^2) \, dy.\]

Step 3: Simplify the expression inside the integral

Simplify the expression inside the integral: \[(3y)^2 - (y^2)^2 = 9y^2 - y^4.\]Thus, the integral looks like:\[V = \pi \int_0^3 (9y^2 - y^4) \, dy.\]

Step 4: Integrate the expression

Integrate the expression for the volume.Find the antiderivative of \(9y^2 - y^4\):\[\int (9y^2 - y^4) dy = 9 \cdot \frac{y^3}{3} - \frac{y^5}{5} = 3y^3 - \frac{y^5}{5}.\]

Step 5: Evaluate the definite integral

Evaluate the definite integral from \(y = 0\) to \(y = 3\) using the antiderivative:\[V = \pi \left[ 3y^3 - \frac{y^5}{5} \right]_0^3.\]Calculate:\[V = \pi \left[ \left(3(3)^3 - \frac{(3)^5}{5}\right) - \left(3(0)^3 - \frac{(0)^5}{5}\right) \right]\]\[= \pi \left[ 81 - \frac{243}{5} \right]\]

Step 6: Simplify and compute the final volume

Simplify:\[81 - \frac{243}{5} = \frac{405}{5} - \frac{243}{5} = \frac{162}{5}.\]Thus, the volume is:\[V = \pi \times \frac{162}{5} = \frac{162\pi}{5}.\]

Key Concepts

Washer Method

The washer method is a popular technique in calculus used to calculate the volume of a solid obtained by rotating a region around an axis. To visualize this, imagine stacking thin washers, which are essentially flat rings. These washers have varying radii, depending on where you are in the stack, which corresponds to the bounded region.

One identifies the outer and inner radii of each washer, determined by the functions that define the boundaries of the solid. The key formula used in the washer method for solids of revolution about the y-axis is:

• Volume, \( V = \pi \int_a^b \left( R^2(y) - r^2(y) \right) \, dy \)

Here, \( R(y) \) is the outer radius, and \( r(y) \) is the inner radius. The integration bounds \( a \) and \( b \) are the intersections of the curves along the y-axis. In our exercise, the outer and inner radii are defined as \( 3y \) and \( y^2 \), respectively. Understanding this method is crucial for calculating volumes of more complex shapes.

Integration

Integration is a fundamental concept in calculus, acting as a way to find accumulations such as areas under curves or volumes of solids. When discussing integration in terms of solids, it allows us to "sum" infinitely small parts to find the total volume. Each infinitesimally small section contributes to the final volume when summed integrally.

The integral takes an expression (often a function), and processes it over a specific range defined by limits. In the case of the washer method, the integration enables us to add up all those small washer volumes to find the total solid volume.
For example, the integral \( \int_0^3 (9y^2 - y^4) \, dy \) calculates the sum of the volumes of the washers created between \( y = 0 \) and \( y = 3 \). This is a perfect example of how integration is used to solve real-world engineering and physics problems where volume calculation is required.

Definite Integral

A definite integral allows you to compute the total accumulation of quality, such as volume or area, between specific limits. Unlike indefinite integrals, which yield a family of functions, definite integrals provide a number as a result.

In our given problem, we calculate the volume of the solid by performing a definite integral from \( y = 0 \) to \( y = 3 \). This process is denoted by \( \left[ f(y) \right]_a^b \), where \( f(y) \) is the antiderivative of the given function.

• The definite integral \( \pi \left[ 3y^3 - \frac{y^5}{5} \right]_0^3 \) gives us the total volume after evaluating it from 0 to 3.

By calculating a definite integral, we consider the specific interval of rotation and thus determine the solid's total volume. This concept is key to effectively using integration for practical applications.

Antiderivative

An antiderivative is essentially the reverse of differentiation. It plays a crucial role in integration, as it helps in finding the actual function or expression over which we are integrating.

The process of finding an antiderivative is known as anti-differentiation. It involves working backwards from a derivative to find the original function. For some functions, finding the antiderivative is straightforward, while for others, it can be more complex.
In our solution, the antiderivatives for the terms \( 9y^2 \) and \( y^4 \) were \( 3y^3 \) and \( \frac{y^5}{5} \), respectively:

• The function \( 9y^2 \) integrates to \( 3y^3 \)
• The function \( y^4 \) integrates to \( \frac{y^5}{5} \)

Using these antiderivatives, we are then able to evaluate the definite integral, and ultimately solve for the volume of the solid. Knowing how to find antiderivatives aids in solving a vast array of mathematical problems.