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Chapter 15: Problem 52

Explain how double integrals can be used to compute the volume of solids in 3 -space.

Short Answer

Expert verified
Double integrals can be used to compute the volume of solids in 3-space by integrating a function f(x,y) that represents the height of the solid with respect to the x-y plane over a specified region of integration. After determining the function and region of integration, set up the double integral with appropriate limits for x and y, then evaluate the integral by first integrating with respect to y and following with the integration with respect to x. The resulting value represents the volume of the solid in cubic units. For example, for a solid with height function \(f(x,y) = 9 - x^2 - y^2\) over a circle centered at the origin with radius 3, the volume is \(\frac{6561\pi}{35}\) cubic units.

Step by step solution

01

Understand Double Integrals

A double integral is a method of integration in which we integrate a function of two variables with respect to one variable and then integrate the resulting function with respect to the other variable. Essentially, it allows us to compute the net accumulation of a function over a two-dimensional region. In the context of finding the volume of a solid in 3-space, we will be integrating a function f(x,y) over a given double integral represents the "infinitesimal volume" of the solid, and the outer integral adds up these infinitesimal volumes over the entire region of interest.
02

Setting up the Double Integral

To set up a double integral for calculating the volume of a solid in 3-space, we need to determine the following: 1. The function f(x,y) representing the height of the solid with respect to the x-y plane. 2. The region of integration, which is the projection of the solid onto the x-y plane. For example, let's consider a solid whose height is given by the function \(f(x,y) = 9 - x^2 - y^2\) and the region of integration is the circle centered at the origin with radius 3.
03

Compute the Limits of Integration

In order to set up the double integral, we need to express the region of integration in terms of the limits for the variables x and y. Given that the region of integration is a circle centered at the origin with radius 3, we can represent the limits of integration as follows: For x: -3 to 3 For y: \(-\sqrt{9 - x^2}\) to \(\sqrt{9 - x^2}\) With these limits, we can set up the double integral for the volume of the solid: \[V = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (9 - x^2 - y^2) \,dy \,dx\]
04

Evaluate the Double Integral

Now that we have the double integral set up, we can evaluate it to find the volume of the solid. First, we will integrate with respect to y, and then integrate the resulting function with respect to x: \[V = \int_{-3}^3 \left[ \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (9 - x^2 - y^2) \,dy \right] \,dx\] After integrating with respect to y, we get: \[V = \int_{-3}^3 \left[-\frac{1}{3}(9-x^2)(9-x^2) + \frac{1}{3}(9-x^2) \cdot 2\sqrt{9-x^2}\right] \, dx\] Then we can integrate with respect to x: \[V = \frac{6561\pi}{35}\] Therefore, the volume of the solid is \(\frac{6561\pi}{35}\) cubic units.

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Key Concepts

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

Volume of Solids
Understanding how to calculate the volume of a solid using double integrals can be intriguing yet straightforward. When dealing with solids in the 3D space, we often want to find how much space a solid occupies. Instead of adding up a bazillion tiny cubes within the solid, we use double integrals to make the process faster and more precise.
This involves the mathematical technique of integration where we deconstruct the solid into infinitesimally small volumes and then sum them back up. Here's the scoop:
  • The function we integrate, often denoted as \(f(x, y)\), represents the height, or thickness, of the solid at any point above the x-y plane. It's like getting little vertical slices of your solid and stacking them along the curve of \(f(x, y)\).
  • Once we set up our integration on a given region, the double integral essentially multiplies this height by the area of the infinitesimal slice and adds all these tiny 3D pixel-like pieces to give us the total volume.
In essence, double integrals help us calculate the volume by piling up these tiny vertical slices over a specified region without the mess of assembling tiny cubes.
3-Space
3-space, also known as three-dimensional space, is where all the action happens for double integrals when calculating volumes. If you think of a regular piece of paper as 2-space, then 3-space is like climbing into that paper and adding another dimension you can move in: up or down.
In our case, 3-space is where the solid exists, projected over an area in the x-y plane.
  • The surface you're integrating over is called the x-y plane, literally a blanket laid out flat, which holds the key to the base area of your solid.
  • Here in 3-space, every point can be described not just by its position (x, y) on this blanket, but also by its height value \(z = f(x, y)\). This is what give shapes life in space, like a peaked mountain range or a mellow hill.
Double integration, therefore, helps us traverse across this 3-space, allowing us to consider the elevation (z) of our solids effectively, resulting in an accurate computation of their volumes.
Integration Limits
The integration limits play a crucial role in setting up a double integral. They define the boundaries, or edges of our solid that we are interested in finding the volume for.
Imagine drawing a closed shape on your x-y plane; this closed shape captures the area beneath which our solid resides. These drawn-out boundaries, expressed as integration limits, tell us where to dig in or where to stop scraping off heights.
  • In practice, you're setting these limits for both x and y coordinates to cover the entire solid's base region, like laying a grid of lines to parcel off sections of your map that you're going to explore.
  • For an example, consider a circle on the x-y plane which would convert to polar form limits. However, if you use Cartesian limits, you'd represent it with ranges like from \(-3\) to \(3\) for \(x\), and from \(-\sqrt{9-x^2}\) to \(\sqrt{9-x^2}\) for \(y\), assuming a radius of 3.
These limits ensure we're not mistakenly computing a non-existent volume by defining precise pathways within our paper map that are too curious to probe beyond and into a new world of 3-space solid objects.

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