Chapter 14: Problem 78
Find the volume of the region that lies inside the sphere \(x^{2}+y^{2}+z^{2}=2\) and outside the cylinder \(x^{2}+y^{2}=1\)
Short Answer
Expert verified
The volume is \( \frac{5\pi\sqrt{2}}{3} \).
Step by step solution
01
Understand the Problem
We need to find the volume of the region that is inside the sphere described by the equation \(x^2 + y^2 + z^2 = 2\) and outside the cylinder described by \(x^2 + y^2 = 1\).
02
Visualize and Set up the Integration Limits
The sphere has a radius of \(\sqrt{2}\), and the cylinder has a radius of 1 in the xy-plane and extends infinitely in the z-direction. We'll use cylindrical coordinates \((r, \theta, z)\), where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). Here, \(r\) ranges from 1 to \(\sqrt{2 - z^2}\).
03
Volume Element in Cylindrical Coordinates
In cylindrical coordinates, the volume element is \(dV = r \, dr \, d\theta \, dz\). We'll integrate \(r\) from 1 to \(\sqrt{2 - z^2}\), \(\theta\) from 0 to \(2\pi\), and \(z\) from \(-\sqrt{2}\) to \(\sqrt{2}\).
04
Set up the Triple Integral
The integral to compute the volume \(V\) is given by the triple integral: \[ V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{0}^{2\pi} \int_{1}^{\sqrt{2 - z^2}} r \, dr \, d\theta \, dz \]
05
Integrate with respect to \(r\)
Perform the integral with respect to \(r\): \[ \int_{1}^{\sqrt{2 - z^2}} r \, dr = \left[ \frac{r^2}{2} \right]_{1}^{\sqrt{2 - z^2}} = \frac{2 - z^2}{2} - \frac{1}{2} \]
06
Integrate with respect to \(\theta\)
Since the integrand after integrating over \(r\) is independent of \(\theta\), the integral over \(\theta\) is simply \[ \int_{0}^{2\pi} 1 \, d\theta = 2\pi \]
07
Integrate with respect to \(z\)
Substitute the results into the remaining integral \[ V = 2\pi \int_{-\sqrt{2}}^{\sqrt{2}} \left( \frac{2 - z^2 - 1}{2} \right) \, dz = \pi \int_{-\sqrt{2}}^{\sqrt{2}} (1 - \frac{z^2}{2}) \, dz \] Integrate with respect to \(z\): \[ \pi \left[ z - \frac{z^3}{6} \right]_{-\sqrt{2}}^{\sqrt{2}} \]
08
Evaluate the Final Integral
Evaluate at \(\sqrt{2}\) and \(-\sqrt{2}\): \[ \pi \left( (\sqrt{2} - \frac{(\sqrt{2})^3}{6}) - (-\sqrt{2} + \frac{(-\sqrt{2})^3}{6}) \right) = \pi \left( 2\sqrt{2} - \frac{(2\sqrt{2})}{6} \right) = \pi \left( 2\sqrt{2} - \frac{\sqrt{2}}{3} \right) \]Simplify: \[ V = \frac{5\pi\sqrt{2}}{3} \]
09
Conclusion
The volume of the region that lies inside the sphere and outside the cylinder is \( \frac{5\pi\sqrt{2}}{3} \).
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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 of expressing points in a space using a combination of a radial distance, an angular coordinate, and a vertical position. This system is particularly useful when dealing with problems involving objects with circular symmetry, like cylinders. In cylindrical coordinates, a point in three-dimensional space is defined as
- \(r\): the radial distance from the z-axis.
- \(\theta\): the angle measured counterclockwise from the positive x-axis.
- \(z\): the same vertical distance as in Cartesian coordinates.
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(z = z\)
Volume Calculation
The calculation of volume using integrals involves determining how much space an object occupies. In our problem, we're using triple integration, which extends the concept of a double integral to three dimensions. For volume calculations in cylindrical coordinates, the volume element is expressed as \(dV = r \, dr \, d\theta \, dz\). Here:
- \(r \, dr\) represents the radial portion of a doughnut-like ring or circular disk.
- \(d\theta\) describes how that ring extends around the z-axis, covering an entire circle.
- \(dz\) accounts for how the space extends vertically from the base to the top of the figure.
- The radial distance \(r\) extends from 1 to \(\sqrt{2 - z^2}\), enclosing the region outside the cylinder.
- \(\theta\) covers 360 degrees, or \(0\) to \(2\pi\), full circle around the z-axis.
- \(z\) stretches from \(-\sqrt{2}\) to \(\sqrt{2}\), encompassing the full vertical extension of the sphere.
Sphere and Cylinder Intersection
Understanding the intersection between a sphere and a cylinder is central to solving the problem. A sphere is defined by its radius and center point, given in the equation form \(x^2 + y^2 + z^2 = R^2\). A cylinder stretches infinitely along one axis, here the z-axis, and is defined by the equation \(x^2 + y^2 = a^2\) where \(a\) is the cylinder's radius.
In our problem, the sphere's equation \(x^2 + y^2 + z^2 = 2\) means a radius of \(\sqrt{2}\). The cylinder is defined as \(x^2 + y^2 = 1\), having a radius of 1.
When these two surfaces intersect, any volume calculations must address how they overlap.
In our problem, the sphere's equation \(x^2 + y^2 + z^2 = 2\) means a radius of \(\sqrt{2}\). The cylinder is defined as \(x^2 + y^2 = 1\), having a radius of 1.
When these two surfaces intersect, any volume calculations must address how they overlap.
- The sphere, with a larger radius, completely encloses the cylinder's cross-section at any given point along the z-axis.
- The volume of interest lies between the cylinder's outer surface (\(r = 1\)) and the sphere's outer boundary (\(r = \sqrt{2 - z^2}\)).