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Magazine Chapter 14: Problem 42
Find the volume \(V\) of the region, using the methods of this section. The region bounded above by the sphere \(x^{2}+y^{2}+z^{2}=1\) and below by the \(x y\) plane
Short Answer
Expert verified
The volume of the hemisphere is \(\frac{2\pi}{3}\).
Step by step solution
01
Understanding the Problem
We are asked to find the volume of a region that is bounded above by a sphere defined by the equation \(x^2 + y^2 + z^2 = 1\) and below by the \(xy\)-plane. This means we are looking for the volume of the upper hemisphere above the \(xy\)-plane.
02
Identifying the Limits of Integration
The sphere \(x^2 + y^2 + z^2 = 1\) is centered at the origin with radius 1. For the hemisphere above the \(xy\)-plane, \(z\) ranges from 0 to \(\sqrt{1 - x^2 - y^2}\). The \(xy\) limits should cover the entire circular base, \(x^2 + y^2 \leq 1\).
03
Setting Up the Integral
The integral to find the volume \(V\) of the hemisphere can be set as:\[V = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} dz \ dy \ dx\]We will integrate with respect to \(z\) first, then \(y\), and finally \(x\).
04
Evaluating the Integral over \(z\)
Integrate with respect to \(z\):\[\int_{0}^{\sqrt{1-x^2-y^2}} 1 \, dz = \sqrt{1 - x^2 - y^2}\]
05
Evaluating the Integral over \(y\)
Substitute the result from the \(z\)-integral into the \(y\)-integral and evaluate:\[\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2} \, dy\]Due to symmetry, this can be rewritten and evaluated more simply using trigonometric substitution.
06
Converting to Polar Coordinates
To evaluate the integral over a circular region in the \(xy\)-plane, convert to polar coordinates where \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). The volume integral becomes:\[V = \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^2} \cdot r \, dr \, d\theta\]
07
Evaluating the Integral in Polar Form
First, evaluate the integral over \(r\):\[\int_{0}^{1} r\sqrt{1-r^2} \, dr = \text{Use substitution } u = 1-r^2\]Then, compute the integral with respect to \(\theta\) over \(0\) to \(2\pi\).
08
Calculating the Final Volume
The detailed evaluation of the integral gives:\[V = \pi \left(\frac{1}{3}\right) = \frac{2\pi}{3}\]The volume of the hemisphere is \(\frac{2\pi}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sphere
A sphere is a three-dimensional object where every point on its surface is equidistant from its center. It can be visualized as a perfectly round ball. The equation of a sphere in three-dimensional Cartesian coordinates is often given by
In the context of the given problem, we consider a sphere with radius 1 centered at the origin. This can be expressed as:
- \( x^2 + y^2 + z^2 = r^2 \)
In the context of the given problem, we consider a sphere with radius 1 centered at the origin. This can be expressed as:
- \( x^2 + y^2 + z^2 = 1 \)
Hemisphere
A hemisphere is half of a sphere, typically divided by a plane passing through its center. This means that a sphere can have two hemispheres:
Visualizing this, the hemisphere looks like a domed cap of the sphere, its base resting on the \(xy\)-plane.
- Northern Hemisphere: The half above the division plane.
- Southern Hemisphere: The half below the division plane.
Visualizing this, the hemisphere looks like a domed cap of the sphere, its base resting on the \(xy\)-plane.
Polar Coordinates
Polar coordinates offer a convenient way to describe the circular base of the hemisphere in the \(xy\)-plane. Instead of using rectangular coordinates \((x, y)\), we use:
This makes our integral setup simpler, paving the way for easier volume calculation.
- \( r \): distance from the origin to a point.
- \( \theta \): angle from the positive \(x\)-axis.
- \( x = r\cos(\theta) \)
- \( y = r\sin(\theta) \)
This makes our integral setup simpler, paving the way for easier volume calculation.
Volume Integral
A volume integral is an extension of basic integration to find the volume under a surface or within a region in three-dimensional space. To solve the problem, we need to set up a triple integral due to the involvement of three variables: \(x\), \(y\), and \(z\).
The integral for the hemisphere can be constructed as follows, initially in Cartesian coordinates:
The integral for the hemisphere can be constructed as follows, initially in Cartesian coordinates:
- \[ V = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} dz \, dy \, dx \]
- \[ V = \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^2} \cdot r \, dr \, d\theta \]
Integration Limits
Integration limits define the boundaries within which the integration occurs. In problems involving geometry, such as calculating the volume of a hemisphere, determining the limits of integration is crucial.
For the given problem, the process involves setting precise limits:
For the given problem, the process involves setting precise limits:
- For \(z\): The upper bound is the surface of the sphere \(z = \sqrt{1-x^2-y^2}\), while the lower bound is the \(xy\)-plane, i.e., \(z = 0\).
- For \(y\): Within a slice of the sphere, the limits range from \(-\sqrt{1-x^2}\) to \(\sqrt{1-x^2}\).
- For \(x\): The integration spans from \(-1\) to \(1\).
- The radius \(r\) ranges from 0 to 1.
- The angle \(\theta\) spans from 0 to \(2\pi\).
