Question

Use cylindrical coordinates to find the volume of the following solid regions. The region in the first octant bounded by the cylinder \(r=1,\) and the planes \(z=x\) and \(z=0\)

Short answer

1. Understand the given region and convert the bounding surfaces to cylindrical coordinates. 2. Determine the limits of integration for r, θ, and z. 3. Set up the volume integral in cylindrical coordinates. 4. Evaluate the integral and simplify to find the volume of the region.

Step-by-step solution

Understanding the given region

The region is bounded by the cylinder \(r=1\), and the planes \(z=x\) and \(z=0\). We are going to find the volume of this region using cylindrical coordinates. In cylindrical coordinates, we have \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and \(z = z\).

Finding the limits of integration

To find the volume of the region, we need to determine the limits of integration for \(r\), \(\theta\), and \(z\). - \(r\) will start from 0 at the origin and end at the cylinder \(r=1\), so we have \(0 \le r \le 1\). - The first octant corresponds to \(\theta\) limiting to the positive x and y values, so \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\), we have \(0 \le \theta \le \frac{\pi}{2}\). - For \(z\), it ranges from the plane \(z=0\) and the plane \(z=x\). In cylindrical coordinates the plane \(z=x\) becomes: \(z=r\cos(\theta)\). So, the limits for \(z\) are \(0 \le z \le r\cos(\theta)\).

Setting up the integral in cylindrical coordinates

We are going to set up the integral in cylindrical coordinates to find the volume of the region, The volume element in cylindrical coordinates is \(dV = r\,dz\,dr\,d\theta\). So the volume integral becomes: $$V = \int_0^{\frac{\pi}{2}} \int_0^1 \int_0^{r\cos(\theta)} r\,dz\,dr\,d\theta$$

Evaluating the integral

We can now evaluate the integral in the following order: \(dz\), \(dr\), and \(d\theta\). $$V = \int_0^{\frac{\pi}{2}} \int_0^1 \left[ \frac{1}{2}r^2\cos(\theta) \right]_0^{r\cos(\theta)} dr\,d\theta$$ $$V = \int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{2}r^3\cos^2(\theta)\,dr\,d\theta$$ Now, we integrate with respect to \(r\): $$V = \int_0^{\frac{\pi}{2}} \left[\frac{1}{8}r^4\cos^2(\theta)\right]_0^1 d\theta$$ $$V = \int_0^{\frac{\pi}{2}}\frac{1}{8}\cos^2(\theta) d\theta$$ Now, we integrate with respect to \(\theta\) using the double-angle formula, \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\): $$V = \frac{1}{8} \int_0^{\frac{\pi}{2}} \frac{1+\cos(2\theta)}{2} d\theta$$ $$V = \frac{1}{16} \left[\theta + \frac{1}{2}\sin(2\theta) \right]_0^{\frac{\pi}{2}}$$ Evaluating the integral and simplifying, we get the volume of the region: $$V = \frac{1}{16}\left[\frac{\pi}{2}\right] = \frac{\pi}{32}$$ Therefore, the volume of the given region in cylindrical coordinates is \(\frac{\pi}{32}\).

Key concepts

Volume of Solid Regions

When studying the volume of solid regions in mathematics, one often encounters situations where converting to a different coordinate system can simplify the process. Cylindrical coordinates are particularly useful when dealing with problems involving circular bounds, such as cylinders or regions that have a natural symmetry around a central axis. The goal is to find the volume of three-dimensional space that a particular solid occupies.

For solids in cylindrical coordinates, the volume can be determined by integrating over the extent of the solid. This involves constructing an integral where each tiny piece, or differential volume element, is summed up over the region. By defining clear boundaries and integrating step by step, the total volume of intricate shapes can be calculated systematically.

Integration Limits

Determining the integration limits in cylindrical coordinates is critical for setting up the integral correctly. For any volume, these limits define the range over which the cylindrical coordinates \(r\), \(\theta\), and \(z\) vary.

In our example, the integration limits are:

• \(r\) spans from 0 to 1 to capture the radial distance from the origin to the boundary of the cylinder.
• \(\theta\) spans from 0 to \(\frac{\pi}{2}\) because we are only considering the first quadrant where both x and y are positive.
• \(z\) ranges from 0 up to \(r\cos(\theta)\), which represents the height limits between the plane \(z=0\) and \(z=x\). This transformed into the cylindrical form relates \(z\) to the radial and angular components.

Correctly defining these limits ensures that the entire solid region is covered without excluding or duplicating any part.

Cylindrical Volume Element

In cylindrical coordinates, the volume element, denoted as \(dV\), represents an infinitesimally small part of the volume. Its form encodes the geometry inherent to cylindrical coordinates. The expression for the cylindrical volume element is \(dV = r\,dz\,dr\,d\theta\).

This element accounts for:

• \(r\), which scales the height and thickness of the element radially.
• \(dz\), the infinitesimal change in height in the direction of the z-axis.
• \(dr\), the radial thickness.
• \(d\theta\), the angular spread which accounts for the arc length in the polar plane.

By integrating this volume element over the specified limits, the entire volume of the region is summed up. It simplifies the calculation of volumes that are symmetrical around the z-axis, making it more manageable than using Cartesian coordinates.

Double-Angle Formula

The double-angle formula is a trigonometric identity often employed when performing integrals involving trigonometric functions. The formula \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\) helps to simplify the integral by transforming a cosine squared function into a more straightforward form.

In our volume integration, the appearance of \(\cos^2(\theta)\) during the integration process makes it ideal to use this formula. By applying the double-angle formula, the integral can be split into more elementary components that are easier to integrate. This transformation turns a potentially complex expression into a linear one involving just cosine. It simplifies the task of finding areas or volumes when trigonometric squares appear, acting as a powerful tool in calculus.