Problem 32 - Calculus: Early Transcendentals

Question

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places. \( y = 0 \) , \( y = \cos^2 x \) , \( \frac{-\pi}{2} \le x \le \frac{\pi}{2} \) (a) About the x-axis (b) About \( y = 1 \)

Step-by-Step Solution

Answer

(a) Volume is approximately 2.46740. (b) Volume is approximately 1.23370.

Step 1: Understanding the Problem (Part a)

We are asked to set up the integral for the volume of the solid obtained by rotating the region bounded by the curves about the x-axis, where the region is described by the curves \( y = 0 \) and \( y = \cos^2 x \) from \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \).

Step 2: Setting Up the Integral (Part a)

For volume calculation by rotation about the x-axis, we use the disk method. The integral is given by \( V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi (\cos^2 x)^2 \, dx \). This represents the volume of disks formed by revolving the region around the x-axis.

Step 3: Understanding the Problem (Part b)

In this part, the region bounded by \( y = 0 \) and \( y = \cos^2 x \) within the same limits is rotated about the line \( y = 1 \). We need to adjust our perspective to account for this axis of rotation, which shifts by 1 unit in the y-direction.

Step 4: Setting Up the Integral (Part b)

For volume calculation by rotation about \( y = 1 \), we use the washer method. The outer radius is \( 1 \) (from \( y = 1 \) to \( y = 0 \)) and the inner radius is \( 1 - \cos^2 x \) (since \( y = 1 \) minus \( y = \cos^2 x \)). The integral is \( V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi [1^2 - (1 - \cos^2 x)^2] \, dx \).

Step 5: Evaluating the Integral (Part a) and (Part b)

Using a calculator, evaluate the integrals derived in the previous steps. For part (a) \( V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^4 x \, dx \), and for part (b) \( V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (2\cos^2 x - \cos^4 x) \, dx \). These integrals yield \( V \approx 2.4674 \) for part (a) and \( V \approx 1.2337 \) for part (b), both to five decimal places.

Key Concepts

Integral Calculus

Integral calculus is a powerful mathematical tool used to find areas, volumes, and other quantities related to accumulation. In the context of finding volumes of solids of revolution, integral calculus helps to compute these volumes when a region in the plane is rotated around a given axis.

The basic idea is to sum up, or integrate, small infinitesimal elements that make up the solid. These elements could be disks, washers, or shells, depending on the method used. Each infinitesimal element can be considered as having a very small volume, and by integrating, we add up these small volumes to find the total volume of the solid.

• The function to be revolved constitutes the boundary of the region.
• The limits of integration correspond to the range over which the region extends.
• The axis of rotation determines the method and form of the integral.

In our exercise, we rotate the function around different axes using the disk and washer methods to calculate the exact volume of the solid.

Disk Method

The disk method is one way to find the volume of a solid of revolution. It is used when a region is revolved around an axis, typically the x-axis or y-axis. In the disk method, the solid is thought of as being made up of numerous thin, flat disks stacked together.

Each disk has a small thickness (say, \(dx\) if rotating around the x-axis) and a radius determined by the function being rotated.

• Formula: \( V = \pi \int [f(x)]^2 \, dx \). Here, \(f(x)\) is the function defining the outer edge of the disk.
• The integral sums the volumes of these disks from the start of the region to the end.
• This method is particularly effective when the axis of rotation is along the boundary of the region (like y=0 in our exercise).

In part (a) of our example, we have a region bounded by \(y = 0\) and \(y = \cos^2 x\). Rotating this about the x-axis, we set up the integral using the disk method to find the volume.

Washer Method

The washer method extends the disk method to situations where there is an inner boundary, creating a hole or 'washer' shape when rotated. This method is notably used when the region is revolved about an axis that is not at the boundary of the region.

In the washer method:

• We calculate the volume of a larger disk (outer radius) and subtract the volume of a smaller disk (inner radius), thus creating a washer.
• Formula: \( V = \pi \int \left( R(x)^2 - r(x)^2 \right) \, dx \).
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