Problem 9 - Thomas Calculus

Question

Two paraboloids Find the volume of the region enclosed by the surfaces \(z=x^{2}+y^{2}\) and \(z=\left(x^{2}+y^{2}+1\right) / 2\)

Step-by-Step Solution

Answer

The volume of the region is \( \frac{\pi}{4} \).

Step 1: Identify the Region of Intersection

The two surfaces in question are paraboloids. The first surface is \( z = x^2 + y^2 \), and the second surface is \( z = \frac{x^2 + y^2 + 1}{2} \). To find the region of intersection, equate the two equations: \( x^2 + y^2 = \frac{x^2 + y^2 + 1}{2} \). Simplifying gives \( 2x^2 + 2y^2 = x^2 + y^2 + 1 \), i.e., \( x^2 + y^2 = 1 \). This is a circle with radius 1 in the \( xy \)-plane.

Step 2: Setup the Integral for Volume Calculation

The volume of interest is the region enclosed between these two surfaces and lying over the circle \( x^2 + y^2 = 1 \). The volume can be computed using the polar coordinate conversion of the difference of the two surfaces. Consider the integral in polar coordinates (\( r, \theta \)): \[ V = \int_{0}^{2\pi} \int_{0}^{1} \left( \frac{r^2 + 1}{2} - r^2 \right) r \, dr \, d\theta \]. Simplifying the integrand: \( \frac{r^2 + 1}{2} - r^2 = \frac{1}{2} - \frac{r^2}{2} \).

Step 3: Evaluate the Integral

Evaluate the integral \( \int_{0}^{1} \left( \frac{1}{2} - \frac{r^2}{2} \right) r \, dr \). This simplifies to \( \int_{0}^{1} \left( \frac{r}{2} - \frac{r^3}{2} \right) \, dr \). Separate into two integrals: \[ \frac{1}{2} \int_{0}^{1} r \, dr - \frac{1}{2} \int_{0}^{1} r^3 \, dr \]. The first integral evaluates to \( \frac{1}{4} \) and the second evaluates to \( \frac{1}{8} \).

Step 4: Compute the Total Volume

The integral over \( r \) gives \( \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \). Now, integrate over \( \theta \): \[ V = \int_{0}^{2\pi} \frac{1}{8} \, d\theta = \frac{1}{8} \times 2\pi = \frac{\pi}{4} \]. This is the total volume of the region between the paraboloids.

Key Concepts

Understanding Paraboloids

Paraboloids are a type of three-dimensional surface that is created by revolving a parabola around its axis. They have a distinct shape like a bowl or an open umbrella. There are two types of paraboloids:

• Elliptic paraboloids, which look like upward or downward facing bowls.
• Hyperbolic paraboloids, which resemble a saddle shape.

In this exercise, we're dealing with elliptic paraboloids described by the equations \( z = x^2 + y^2 \) and \( z = \frac{x^2 + y^2 + 1}{2} \). These paraboloids open upwards in the \(z\)-direction. To solve this problem, we need to understand where these surfaces intersect in the 3D space.
The intersection forms a circular region in the \( xy \)-plane, identified by setting the two equations equal to each other. In this case, it simplifies to the circle equation \( x^2 + y^2 = 1 \). Understanding the geometry of paraboloids is crucial to visualizing and solving problems related to the volume they enclose.

Introduction to Polar Coordinates

Polar coordinates offer a convenient way to define locations on a plane using an angle and a distance from a central point (usually the origin). The coordinates are denoted by \( (r, \theta) \), where \( r \) is the radial distance and \( \theta \) is the angle measured in radians from a reference direction.

• They are particularly useful in problems involving symmetry around a point, like circles.
• This system simplifies integration in circular regions.

In this volume calculation exercise, we transform the Cartesian coordinates \((x, y)\) into polar coordinates because the region of interest is a circle centered at the origin. This switch converts our domain and transforms \( x^2 + y^2 \) into \( r^2 \), which dramatically simplifies the integrals needed for volume calculations.

Basics of Integration

Integration is a fundamental concept used to find areas, volumes, and other quantities. It is the process of finding the integral of a function, which can be thought of as the mathematical counterpart of the sum. In volume calculation, integration helps us sum up infinitesimal bits of volume to get a total volume.

• Double integration is used to find volumes under surfaces over certain domains.
• In polar coordinates, the orders of integration are typically \( r \) and then \( \theta \).

In this exercise, the volume is calculated by setting up an integral over the circular region of intersection. The integral accounts for the difference between the paraboloid surfaces. It transforms into a product of two integrals: one integrating in the radial direction \(r\), and the other integrating in the angular direction \(\theta\). Evaluating these integrals gives us the enclosed volume.

Computing the Volume of Solids

When computing the volume of solids among different surfaces, we consider the area between them within specified boundaries. This task usually involves setting up and evaluating integrals over the domain, often using symmetry or other properties to simplify the computation.
In the given problem, the solid is bounded by two paraboloids, forming a region that is effectively capped at the top and bottom. By computing the volume of such a region, we look at how much 'space' is enclosed.

• The volume between the paraboloids is found by subtracting one surface equation from the other and integrating over the given region.
• Integration in terms of polar coordinates further streamlines this process given the circular boundary.

The concept demonstrated here shows how integration can compute complex volumes, offering insight into methods of approaching geometrical problems that might otherwise seem unwieldy.