Problem 21 - 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. $$ y=x^{2}, x=0, x=3 $$

Step-by-Step Solution

Answer

The volume is \(40.5\pi\) cubic units.

Step 1: Identify the Bounded Area

The region is bounded by the parabola \(y = x^2\), the vertical line \(x = 0\), and \(x = 3\). On the \(xy\) plane, this creates a vertically-oriented region from \(x = 0\) to \(x = 3\).

Step 2: Set Up the Integral Using the Washer Method

The washer method is used when revolving around the \(y\)-axis. The volume formula is \( V = \pi \int_{a}^{b} [R(y)]^2 \, dy \). Since we already have \(y = x^2\), we need \(x\) in terms of \(y\), which gives \(x = \sqrt{y}\). The integral becomes \( V = \pi \int_{0}^{9} [3]^2 - [\sqrt{y}]^2 \, dy \).

Step 3: Determine the Integral Bounds

The lowest value of \(y\) when \(x = 0\) is \(y = 0\), and the highest value of \(y\) when \(x = 3\) is \(y = 9\). Thus, our integral should be evaluated from \(y = 0\) to \(y = 9\).

Step 4: Simplify the Expression Inside the Integral

Substituting the expressions into the integral, \([3]^2 = 9\) and \([\sqrt{y}]^2 = y\). Thus, the integral becomes \( V = \pi \int_{0}^{9} (9 - y) \, dy \).

Step 5: Evaluate the Integral

Evaluate the integral \( \int_{0}^{9} (9 - y) \, dy \). The antiderivative of \(9\) is \(9y\) and the antiderivative of \(y\) is \(\frac{y^2}{2}\). This gives us \(9y - \frac{y^2}{2}\).

Step 6: Calculate the Definite Integral

The definite integral is \([9y - \frac{y^2}{2}]_{0}^{9}\). First, substitute \(y = 9\) to get \(9(9) - \frac{9^2}{2} = 81 - 40.5 = 40.5\). Then substitute \(y = 0\) which gives \(0 - 0 = 0\).

Step 7: Compute the Final Volume

Substitute the values into the volume formula: \( V = \pi [40.5 - 0] = 40.5\pi \). So the volume is \(40.5\pi\) cubic units.

Key Concepts

Washer Method

The Washer Method is a technique used to find the volume of a solid of revolution, particularly when the solid is created by revolving an area around a specific axis. It works by imagining the area as a series of thin washers or disks, with a hole in the middle, stacked along the axis of rotation.

To apply the Washer Method, follow these steps:

• Identify the outer and inner radius of each washer, which represent the distance from the axis of rotation to the outer and inner curves, respectively.
• Set up the integral to calculate the volume, using the formula: \[V = \pi \int_{a}^{b} \left([R(y)]^2 - [r(y)]^2\right) \, dy\]

In this exercise, the region is bounded by the parabola and vertical lines, and rotated around the y-axis. Here, our outer radius is constant (3 units) because it's the distance between the axis of rotation and the line at x = 3, and the inner radius is given by \( \sqrt{y} \), because this is the function x in terms of y.

Definite Integral

A definite integral is a mathematical concept that represents the accumulation of quantities, such as area, under a curve within specific bounds. It is used extensively in calculating volumes, areas, and in other applications of calculus.

In the context of finding volumes, a definite integral helps in summing up the infinitesimally small volumes made by the washers as you stack them along the axis. For example, when revolving the given area around the y-axis, we compute:

• The integral \[\pi \int_{0}^{9} (9 - y) \, dy\]provides the total volume by calculating the difference between the outer and inner radii squared, summed over the interval from y = 0 to y = 9.

The bounds of the integral reflect the height of the region being revolved.

Antiderivative

In calculus, finding an antiderivative is the reverse process of differentiation. It's essential for solving definite integrals, and it essentially answers the question: "What function had a given function as its derivative?"

When calculating the volume using the Washer Method, you will need to integrate functions like constant values and polynomials. The antiderivative of constant 9 with respect to y is 9y, while the antiderivative of y is \( \frac{y^2}{2} \). This means, in our volume problem:

• The expression inside the integral, \(9 - y\),is simplified using their respective antiderivatives: \(9y - \frac{y^2}{2}\).

This allows you to evaluate the integral over the specified bounds to compute the total volume.

Rotating Around y-axis

When a region is rotated around the y-axis, it creates a solid of revolution. This rotation turns a flat, two-dimensional area into a three-dimensional object by spinning it around the vertical y-axis, much like turning a piece of paper into a lampshade by wrapping it around a stick.

In our example, we take the area between the curve \(y = x^2\) and the lines \(x = 0\) and \(x = 3\). By rotating this around the y-axis, each x-value corresponds to circular cross-sections or washers, where the difference in radii forms the hollow part of the washer.

The radial distance from the y-axis each point is converted into a function of y, turning the problem of finding areas into a more manageable problem of finding distances from curves plotted against y values. This approach is pivotal for reaching accurate volume calculations in problems involving rotations around the y-axis.