Question

Find the volume of the solid obtained by revolving about the \(y\) -axis the region bounded by \(x=y^{2}+1, x=0, y=-1,\) and \(y=1\).

Short answer

Answer: The volume of the solid is \(\frac{16\pi}{15}\).

Step-by-step solution

Draw and identify the region

Sketch the functions x = y^2 + 1, x = 0, y = -1, and y = 1 to visualize the region that is being revolved around the y-axis. You can see that the region is a vertical strip between the y function and the y-axis, where -1 ≤ y ≤ 1.

Radius of the disk

In order to apply the disk method, we need to express the radius of the disk as a function of y. In this case, the radius of the disk will be the distance from the y-axis to the function x = y^2 + 1. Since the function is already in the form x = f(y), the radius is just r(y) = y^2 + 1.

Set up the integral

The volume of the solid obtained by revolving the region about the y-axis is given by the integral of the area of the disks from the lower bound to the upper bound. The area of each disk is given by pi times the square of the radius. We will integrate this area with respect to y from -1 to 1. The integral used to find the volume of the solid will be: \[V = \int_{-1}^1 \pi r(y)^2 dy\]

Plug in the radius function

Now, we can plug in our radius function, y^2 + 1, into the integral: \[V = \int_{-1}^1 \pi (y^2 + 1)^2 dy\]

Integrate

To find the volume, we need to evaluate the integral: \[V = \int_{-1}^1 \pi (y^2 + 1)^2 dy = \pi \int_{-1}^1 (y^4 + 2y^2 + 1) dy\] \[V = \pi \left[\frac{y^5}{5} + \frac{2y^3}{3} + y\right]_{-1}^1\]

Evaluate the integral and find the volume

Lastly, we will evaluate the integral at the bounds and subtract to find the volume: \[V = \pi \left[\left(\frac{1^5}{5} + \frac{2(1)^3}{3} + 1\right) - \left(\frac{(-1)^5}{5} + \frac{2(-1)^3}{3} - 1\right)\right]\] \[V = \pi \left[\left(\frac{1}{5} + \frac{2}{3} + 1\right) - \left(-\frac{1}{5} - \frac{2}{3} - 1\right)\right]\] \[V = \pi \left[2\left(\frac{1}{5} + \frac{2}{3} + 1\right)\right]\] \[V = 2\pi\left(\frac{1}{5} + \frac{2}{3} + 1\right) = 2\pi\left(\frac{8}{15}\right) = \frac{16\pi}{15}\] The volume of the solid obtained by revolving the region about the y-axis is \(V = \frac{16\pi}{15}\).

Key concepts

Disk Method

The disk method is a powerful technique used in calculus to find the volume of a solid formed by revolving a region around a given axis. The idea is simple: break down the solid into a series of thin, flat disks (or washers) that are perpendicular to the axis of revolution.

Imagine slicing a loaf of bread, where each slice represents a disk. The cumulative volume of these disks is approximately the volume of the entire solid. In our scenario, we're revolving the region bounded by these limits around the y-axis.

**How to Use the Disk Method**
To apply the disk method efficiently, follow these steps:

• Identify the region being revolved.
• Set up the radius of a typical disk, as a function of the variable of integration.
• Use the formula for the volume of a disk: \( V = \pi \int [r(y)]^2 dy \), where \([r(y)]\) is the radius.
• Evaluate the definite integral to find the total volume.

Definite Integral

The definite integral plays a crucial role in calculating the volume of solids. It helps sum up the infinitely many, infinitely small volumes of the disks to get the overall volume of the solid.

When working with the disk method, the definite integral is set up from one boundary of the region to the other. This ensures that all disks from the bottom to the top of the solid are included.

**Integral Setup and Properties**

• The integral is often presented in the form \( \int_a^b f(y) dy \), where \( a \) and \( b \) are the limits of integration representing the boundaries of the region.
• This integral computes the net area under the curve, which in the case of volumes, translates into the net volume of the solid.

Once you evaluate this integral, you get a precise measurement of the volume.

Revolution around y-axis

Revolution around the y-axis is a common practice in problems involving volumes of solids. This concept revolves around rotating a region about the vertical y-axis to form a three-dimensional object.

For our problem, the curve delimiting the region to be revolved is defined by the equation \(x = y^2 + 1\). This defines the outer boundary of the disk's radius.

**Understand the Geometric Transformation**

• Consider how a point on the boundary rotates to become part of a circular path.
• Every point along the boundary curve traces out a circle as it spins around the axis, forming the edge of a disk.

This geometric transformation is the heart of what turns a 2D region into a solid 3D volume.

Polynomial Integration

Polynomial integration involves finding the integral of polynomial functions, which is necessary to solve many volume problems involving the disk method.

In our exercise, the expression for the radius squared, \((y^2 + 1)^2\), expands into a polynomial: \( y^4 + 2y^2 + 1 \). This breakdown is essential for integration.

**Steps for Polynomial Integration**

• Expand the polynomial if necessary, to ease integration.
• Integrate each term separately: \( \int y^n dy \) is computed as \( \frac{y^{n+1}}{n+1} \).
• Evaluate the antiderivative, applying the limits of integration to get the definite result.

Each term’s integral simplifies the process of finding the volume of the solid when bounded by polynomial expressions.