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Chapter 4: Problem 76

For the following problems, find the specified area or volume.The volume of the solid that lies between the paraboloid \(z=2 x^{2}+2 y^{2}\) and the plane \(z=8\).

Short Answer

Expert verified
The volume is \(16\pi\).

Step by step solution

01

Define the Region

The paraboloid given is described by the equation \(z = 2x^{2}+2y^{2}\). To determine the region we are dealing with, note that the solid is bounded by the plane \(z = 8\). The volume we seek lies above the paraboloid and below this plane.
02

Project onto the XY-plane

To find the region of integration, we need to set the two equations \(z = 2x^{2}+2y^{2}\) and \(z = 8\) equal to each other to find where they intersect. Set \(2x^{2}+2y^{2}=8\), which simplifies to \(x^{2} + y^{2} = 4\). This equation represents a circle of radius 2 centered at the origin in the XY-plane.
03

Set Up the Integral

We will use cylindrical coordinates to find the volume. In cylindrical coordinates, the region where the volume lies is a circle of radius 2. Thus, \(r\) goes from 0 to 2, \(\theta\) goes from \(0\) to \(2\pi\), and \(z\) goes from \(2r^{2}\) to 8. The volume integral becomes: \[ V = \int_{0}^{2\pi} \int_{0}^{2} \int_{2r^{2}}^{8} r \, dz \, dr \, d\theta \]
04

Evaluate the Integral

First, evaluate the integral with respect to \(z\):\[ \int_{2r^{2}}^{8} 1 \, dz = [z]_{2r^{2}}^{8} = 8 - 2r^{2} \]Now, evaluate the integral with respect to \(r\): \[ \int_{0}^{2} (8 - 2r^{2})r \, dr = \int_{0}^{2} (8r - 2r^{3}) \, dr \]Integrate term by term:\[ = [4r^{2} - \frac{r^{4}}{2}]_{0}^{2} = [4(4) - \frac{16}{2}] = 16 - 8 = 8 \]Finally, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} 8 \, d\theta = 8 \times (2\pi - 0) = 16\pi \]
05

State the Final Volume

After integrating through all variables, we find the volume of the solid between the paraboloid \(z = 2x^{2} + 2y^{2}\) and the plane \(z = 8\) is \(16\pi\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Coordinates
Cylindrical coordinates are a way to explore three-dimensional space using two-dimensional polar coordinates combined with a height from the plane.
Imagine a cylindrical column where points are identified by:
  • \( r \) - the radial distance from the origin to the point's projection on the xy-plane,
  • \( \theta \) - the angle between the positive x-axis and the point's projection on the xy-plane,, and
  • \( z \) - the height above the xy-plane.
This system is highly beneficial when dealing with circular regions like the one in the exercise. Here, since the solid's projection on the xy-plane is a circle, cylindrical coordinates let us leverage the symmetry, reducing the complexity of setting up and evaluating integrals, especially for volumes of solids.
Double Integration
Double integration is a powerful technique used to compute volumes under surfaces or areas over planes.
In this process, we integrate a function continuously over two variables, typically within a region on the xy-plane.
For this exercise:
  • The first integral is with respect to \( z \), determining the slice of volume above the paraboloid and below the plane.
  • The second integral, over variable \( r \), sums these volumes from the center outwards.
  • Finally, the integration with respect to \( \theta \) accounts for every direction around the circle, finalizing the total volume.
By using bounds defined by the problem, double integration lets us accumulate small volumes into the total volume, following the curve of the paraboloid and the plane's constraints.
Paraboloids
A paraboloid is a three-dimensional surface shaped like an elongated or flattened parabola.
In mathematics, paraboloids come in two main types:
  • Elliptic paraboloids鈥攃urved like an upward or downward bowl.
  • Hyperbolic paraboloids鈥攕addle-shaped.
The solid in the exercise is bounded above by the plane \( z = 8 \) and below by the elliptic paraboloid described by \( z = 2x^2 + 2y^2 \).
The equation of the paraboloid shows symmetry in the \( x \) and \( y \) directions, making it ideal for solving via cylindrical coordinates.This symmetry and its simple intersection with the plane create a manageable region for integration, allowing for straightforward calculation of the precise volume.
Understanding the nature of paraboloids helps in visualizing the solid and setting up integrals that match this symmetry.

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