Problem 54 - University Calculus: Early Transcendentals

Question

Find the volume of the region bounded below by the paraboloid \(z=x^{2}+y^{2},\) laterally by the cylinder \(x^{2}+y^{2}=1,\) and above by the paraboloid \(z=\) \(x^{2}+y^{2}+1\).

Step-by-Step Solution

Answer

The volume is \( \pi \).

Step 1: Identify the boundaries

The given paraboloids are \( z = x^2 + y^2 \) and \( z = x^2 + y^2 + 1 \). The lateral boundary is the cylinder \( x^2 + y^2 = 1 \). Our region of interest lies between the two paraboloids and within this cylinder.

Step 2: Set up the volume integral

We are working with two paraboloids and a cylindrical boundary. We will use cylindrical coordinates because it naturally fits the cylindrical boundary: \( r^2 = x^2 + y^2 \), \( z = r^2 \), and \( z = r^2 + 1 \). The limits for \( r \) will be from 0 to 1, and for \( \theta \) from 0 to \( 2\pi \).

Step 3: Determine the integrand

The integrand is the difference between the two paraboloids since we want the volume between them: \( (r^2 + 1) - r^2 = 1 \). So, we integrate this constant value with respect to the cylindrical coordinates.

Step 4: Write the integral in cylindrical coordinates

The volume is given by the double integral \[ V = \int_{0}^{2\pi} \int_{0}^{1} (1) \, r \, dr \, d\theta \]The extra \( r \) in the integrand comes from converting the Cartesian area element \( dx \, dy \) to the cylindrical area element \( r \, dr \, d\theta \).

Step 5: Integrate with respect to r

Calculate the integral with respect to \( r \):\[ \int_{0}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1} = \frac{1}{2} \]

Step 6: Integrate with respect to \( \theta \)

Now calculate the integral with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times \left[ \theta \right]_{0}^{2\pi} = \frac{1}{2} \times 2\pi = \pi \]

Step 7: Conclusion

The volume enclosed by the two paraboloids and the cylindrical boundary is \( \pi \).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a way to represent points in space using a combination of radial distance, angle, and height. Unlike Cartesian coordinates, which use

• x,
• y,
• z-axis labels,

here we instead use:

• a radial distance (r),
• angle (\(\theta\)),
• and height (z).

Think of this as a natural extension of polar coordinates into three dimensions. In problems with circular symmetry, such as those involving cylinders, these coordinates simplify the math considerably.
To convert from Cartesian to cylindrical coordinates:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)
• \(z = z\)

In the original exercise, we used cylindrical coordinates because the boundary of interest, a cylinder, was inherently circular. This choice allowed for more straightforward integration and simplified calculations. Understanding the importance of choosing the right coordinate system for integration can aid in solving problems more efficiently.

Paraboloid

A paraboloid is a three-dimensional surface that can be visualized as a continuous cross-section of parabolas. It comes in two main types: elliptical and hyperbolic. Our focus here is on the elliptical paraboloid, which, in simple terms, looks like a bowl or dish. This shape is often described using the equation \(z = x^2 + y^2\), which opens upwards.
Let's break down the behavior of the function:

• As you move away from the origin in any radial direction, the z value increases.
• The smallest z value is at the origin: \((x, y) = (0,0)\).
• The resulting surface is smooth and symmetrical around the z-axis.

In the problem at hand, we studied the region between two paraboloids:\( z = x^2 + y^2 \) and \( z = x^2 + y^2 + 1 \). This denotes two parallel paraboloid surfaces. Here, the volume of interest is the space between these two surfaces, achievable because their equations define clear and symmetrical bounds.

Cylinder

A cylinder is a three-dimensional shape defined by its circular base and uniform height. Its cross-section at any parallel level is a circle. In this problem, the cylinder is described by the equation \(x^2 + y^2 = 1\), which represents all points equidistant from the origin, lying on a circle of radius 1 centered at the origin.
This equation defines the lateral boundary for the region we are looking at. Let's understand its key features:

• The cylinder has an infinite length in the z-direction unless otherwise specified.
• In this exercise, our cylinder acts as the side walls demarcating the boundary for the paraboloid surfaces.
• Cylindrical coordinates are particularly useful here because they align perfectly with the circular nature of the boundary.

Because of the symmetry of cylinders, calculus problems involving their volume or regions tend to be easier to navigate and solve when using cylindrical coordinates. The symmetry reduces the problem size and complexity, making it a preferred choice whenever a circular boundary is involved.