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Find the volume enclosed by \(x^{2}+y^{2}+z^{2}=a^{2}\) using (a) cylindrical coordinates (b) spherical coordinates.

Short Answer

Expert verified
The volume of the sphere is \(\frac{4\pi}{3}a^3\) using both cylindrical and spherical coordinates.

Step by step solution

01

(a): Identify the conversion in cylindrical coordinates

Cylindrical coordinates relate to Cartesian coordinates as follows: \[ x = r \cos(\theta), \quad y = r \sin(\theta), \quad z = z \]where \( r \) is the radial distance \( r = \sqrt{x^2 + y^2} \) and \( \theta \) is the azimuthal angle. The equation of the sphere \( x^2 + y^2 + z^2 = a^2 \) can be expressed in cylindrical coordinates as:\[ r^2 + z^2 = a^2 \]
02

(a): Set up the integral for volume in cylindrical coordinates

To find the volume, we integrate the volume element \( dV = r \, dr \, d\theta \, dz \) over the domain described by \( r^2 + z^2 \leq a^2 \). We need to choose bounds for \( z \) and \( \theta \):- \( \theta \) ranges from 0 to \( 2\pi \)- For each \( z \), \( r \) ranges from 0 to \( \sqrt{a^2 - z^2} \)- \( z \) ranges from \( -a \) to \( a \)
03

(a): Evaluate the integral in cylindrical coordinates

The integral for volume in cylindrical coordinates becomes:\[ V = \int_{0}^{2\pi} \int_{-a}^{a} \int_{0}^{\sqrt{a^2 - z^2}} r \, dr \, dz \, d\theta \]Calculate the innermost integral (with respect to \( r \)), the result is:\[ \int_{0}^{\sqrt{a^2 - z^2}} r \, dr = \frac{1}{2}(a^2 - z^2) \]Then, integrate with respect to \( z \):\[ \int_{-a}^{a} \frac{1}{2}(a^2 - z^2) \, dz = \frac{2}{3}a^3 \]Finally, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{2}{3}a^3 \, d\theta = \frac{4\pi}{3}a^3 \]
04

(b): Identify the conversion in spherical coordinates

Spherical coordinates are given by \((\rho, \phi, \theta)\), where:\[ x = \rho \sin(\phi) \cos(\theta), \; y = \rho \sin(\phi) \sin(\theta), \; z = \rho \cos(\phi) \]The equation for the sphere \( x^2 + y^2 + z^2 = a^2 \) becomes \( \rho^2 = a^2 \), thus \( \rho = a \).
05

(b): Set up the integral for volume in spherical coordinates

The volume element in spherical coordinates is \( dV = \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \). Integrate over the sphere defined by \( 0 \leq \rho \leq a \), \( 0 \leq \phi \leq \pi \), and \( 0 \leq \theta \leq 2\pi \).
06

(b): Evaluate the integral in spherical coordinates

Write the volume integral in spherical coordinates:\[ V = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]First, compute the innermost integral (with respect to \( \rho \)):\[ \int_{0}^{a} \rho^2 \, d\rho = \frac{1}{3}a^3 \]Next, integrate with respect to \( \phi \):\[ \int_{0}^{\pi} \sin(\phi) \, d\phi = 2 \]Lastly, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{1}{3}a^3 \, \cdot 2 \, d\theta = \frac{4\pi}{3}a^3 \]
07

Conclusion: Compare both methods

Both (a) cylindrical and (b) spherical coordinate methods give the same result for the volume enclosed by the sphere:\[\frac{4\pi}{3}a^3\]

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Key Concepts

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

Cylindrical Coordinates
Cylindrical coordinates extend the idea of two-dimensional polar coordinates into three dimensions by introducing an additional coordinate for height, often denoted as \( z \). These coordinates are expressed as \((r, \theta, z)\):
  • \( r \) is the radial distance from the origin to the point's projection onto the \( xy \)-plane, calculated as \( r = \sqrt{x^2 + y^2} \).
  • \( \theta \) is the angle between the positive \( x \)-axis and the line from the origin to the projection of the point onto the \( xy \)-plane.
  • \( z \) is the height of the point above or below the \( xy \)-plane, the same as in Cartesian coordinates.
The conversion between Cartesian (\( x, y, z \)) and cylindrical coordinates is:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
  • \( z = z \)
For the given problem of finding the volume of a sphere, transforming into cylindrical coordinates makes sense because the equation of the sphere simplifies to \( r^2 + z^2 = a^2 \), which is easier to work with when setting up and evaluating integrals.
Spherical Coordinates
Spherical coordinates are a system that extends the concept of polar coordinates into three dimensions using three parameters: radius, polar angle, and azimuthal angle. In spherical coordinates, a point is represented by \((\rho, \phi, \theta)\):
  • \( \rho \) is the radial distance from the origin to the point, equivalent to the radius. In the case of a sphere of radius \( a \), \( \rho \) equals \( a \).
  • \( \phi \) is the polar angle measured from the positive \( z \)-axis downwards, ranging from 0 to \( \pi \).
  • \( \theta \) is the azimuthal angle in the \( xy \)-plane, similar to \( \theta \) in cylindrical coordinates, ranging from 0 to \( 2\pi \).
Coordinates are converted as follows:
  • \( x = \rho \sin(\phi) \cos(\theta) \)
  • \( y = \rho \sin(\phi) \sin(\theta) \)
  • \( z = \rho \cos(\phi) \)
In this context, using spherical coordinates to calculate the volume of a sphere is typically more straightforward because the limits of integration are constants and the volume integral simplifies directly to the formula for the volume of a sphere.
Triple Integration
Triple integration is an extension of double integration to find the volume under a surface in three-dimensional space. It calculates the integral over a three-dimensional region, often involving a volume element. For the sphere, the volume element differs based on the coordinate system:
  • In cylindrical coordinates, the volume element is \( dV = r \, dr \, d\theta \, dz \).
  • In spherical coordinates, the volume element is \( dV = \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \).
Each element expresses how volume increments vary based on the system used. When setting up the integral, the limits of integration represent the bounds of the region where the volume is determined. Evaluating triple integrals requires performing integration individually with respect to each variable, from the innermost to the outermost integral. This ensures an accurate computation of volume for the specified region or object. The process allows for practical applications across physics and engineering, such as calculating the volume of complex shapes that cannot be easily measured using simple geometric formulas.

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Most popular questions from this chapter

Evaluate the iterated integral by converting to polar coordinates. $$ \int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} \frac{d y d x}{\left(1+x^{2}+y^{2}\right)^{3 / 2}} \quad(a>0) $$

Use polar coordinates to evaluate the double integral. \(\iint_{R} \sqrt{9-x^{2}-y^{2}} d A,\) where \(R\) is the region in the first quadrant within the circle \(x^{2}+y^{2}=9\)

The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\ I_{z} &=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions in Exercises 45 and 46 Consider the rectangular lamina that occupies the region described by the inequalities \(0 \leq x \leq a\) and \(0 \leq y \leq b\). Assuming that the lamina has constant density \(\delta,\) show that $$I_{x}=\frac{\delta a b^{3}}{3}, \quad I_{y}=\frac{\delta a^{3} b}{3}, \quad I_{z}=\frac{\delta a b\left(a^{2}+b^{2}\right)}{3}$$

The transformation \(x=a u, y=b v(a>0, b>0)\) can be rewritten as \(x / a=u, y / b=v,\) and hence it maps the circular region $$ u^{2}+v^{2} \leq 1 $$ into the elliptical region $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 $$ In these exercises, perform the integration by transforming the elliptical region of integration into a circular region of integration and then evaluating the transformed integral in polar coordinates. \(\iint_{R} \sin \left(4 x^{2}+9 y^{2}\right) d A,\) where \(R\) is the region in the first quadrant enclosed by the ellipse \(4 x^{2}+9 y^{2}=1\) and the coordinate axes.

Use an appropriate change of variables to find the volume of the solid bounded above by the plane \(x+y+z=9,\) below by the \(x y\) -plane, and laterally by the elliptic cylinder \(4 x^{2}+9 y^{2}=36 .\) [Hint: Express the volume as a double integral in \(x y\) -coordinates, then use polar coordinates to evaluate the transformed integral.]

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