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Find the moment of inertia of a right circular cone of base radius 1 and height 1 about an axis through the vertex parallel to the base. (Take \(\delta=1\).)

Short Answer

Expert verified
\( \frac{\pi}{10} \)

Step by step solution

01

Understand the Problem

We are asked to find the moment of inertia of a right circular cone about an axis through the vertex parallel to the base. The cone has a height and base radius of 1, and density of 1.
02

Set up the Integral for the Moment of Inertia

The moment of inertia of a solid about an axis is given by \[ I = \int_V \rho r^2 \, dV \]for a continuous mass distribution, where \( ho \) is the density, and \( r \) is the perpendicular distance from the axis of rotation. Here, \( \rho = 1 \).
03

Identify the Volume Element

For integration in cylindrical coordinates, the volume element is \[ dV = r \, dr \, d\theta \, dz \]where \( r \) is the distance from the axis (vertical through the vertex), \( \theta \) is the angular coordinate, and \( z \) is the height coordinate.
04

Determine Integration Limits

The height \( z \) runs from 0 to 1, \( \theta \) runs from 0 to \( 2\pi \), and the base radius varies linearly with height as \( r = z \) (because the slope of the cone wall from base to vertex is \( 1 \)).
05

Set up the Integral

The integral becomes:\[ I = \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{z} r^2 \cdot r dr \, d\theta \, dz = \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{z} r^3 \, dr \, d\theta \, dz \]
06

Evaluate the Integral with Respect to r

First, integrate with respect to \( r \):\[ \int_{0}^{z} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{z} = \frac{z^4}{4} \]
07

Evaluate the Integral with Respect to \( \theta \)

Next, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{z^4}{4} \, d\theta = \frac{z^4}{4} \times 2\pi = \frac{\pi z^4}{2} \]
08

Evaluate the Integral with Respect to z

Finally, integrate with respect to \( z \):\[ \int_{0}^{1} \frac{\pi z^4}{2} \, dz = \frac{\pi}{2} \int_{0}^{1} z^4 \, dz = \frac{\pi}{2} \left[ \frac{z^5}{5} \right]_{0}^{1} = \frac{\pi}{2} \times \frac{1}{5} = \frac{\pi}{10} \]
09

Conclusion

Thus, the moment of inertia of the cone about the specified axis is \( \frac{\pi}{10} \).

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Key Concepts

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

Cylindrical Coordinates
When dealing with problems involving three-dimensional objects like cones, spherical, and cylindrical shapes, cylindrical coordinates simplify the calculations. Cylindrical coordinates are particularly useful when the object exhibits symmetry around an axis, such as a cone or a cylinder. This coordinate system is defined by three values:
  • The radial distance, \( r \): the distance from the origin to the projection of the point in the \( xy \)-plane. For a cone, this varies linearly as \( r = z \), given a slope of 1 from the base to the vertex.
  • The angle \( \theta \): the component of cylindrical coordinates, representing the angle at which the point lies in the plane. For a complete rotation, it ranges from 0 to \( 2\pi \), encapsulating full circular symmetry.
  • The height, \( z \): this is the vertical height above the \( xy \)-plane, ranging for our cone from 0 to 1, matching the cone's total height.
Cylindrical coordinates make it easy to describe and compute properties of these shapes by capitalizing on their symmetric nature around a central axis.
Integral Calculus
To find the moment of inertia, integral calculus steps in as an invaluable tool by allowing us to sum up infinitesimally small contributions across an entire volume. The moment of inertia is calculated using the integral\[I = \int_V \rho r^2 \, dV\]where \( \rho \) is the density, in this case, 1 for simplicity, and \( r^2 \) represents the squared distance from our chosen axis. For our cone, the differential element of volume (\( dV \)) in cylindrical coordinates becomes \( r \, dr \, d\theta \, dz \), integrating from the base of the cone to the top, and fully around its circle base; \( \theta \) changes from 0 to \( 2\pi \), covering the full circular aspect.The integration proceeds logically:
  • First, with respect to \( r \), the radial coordinate from 0 to \( z \), the height linearly works as a restraining factor of the radius.
  • Then, with respect to the angular coordinate \( \theta \), integrating over a full circle.
  • Lastly, the vertical coordinate \( z \) from base 0 to peak 1.
This sequential integration accounts for every small piece adding up to the moment of inertia for the complete cone.
Solid Geometry
Understanding the geometric properties of solids like cones is crucial in problems regarding moments of inertia. A right circular cone provides a distinct shape defined by its circular base and a vertex directly above its center. For calculating physical properties like mass, volume, or moments of inertia, recognizing:
  • The base radius (1 in our case) defines how broad and circular the structure is. In integration, it directly affects limits and ranges.
  • The height, likewise given as 1, positions the vertex and limits vertical integration.
  • Symmetry about the axis through the vertex parallel to the base simplifies calculations using cylindrical coordinates.
This clear understanding allows us to apply mathematical techniques efficiently, seeing how changes in one dimension influence overall integration and turning abstract calculus operations into tangible geometry understanding.

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Most popular questions from this chapter

Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid that is bounded on the front and back by the planes \(x=2\) and \(x=1,\) on the sides by the cylinders \(y=\pm 1 / x,\) and above and below by the planes \(z=x+1\) and \(z=0\)

In Exercises \(49-52,\) use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region. \(F(x, y, z)=x^{4}+y^{2}+z^{2}\) over the solid sphere \(x^{2}+y^{2}+\) \(z^{2} \leq 1\).

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\)-plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\)-plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility. $$\int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x$$

Find the volume of the smaller region cut from the solid sphere \(\rho \leq 2\) by the plane \(z=1\).

Let \(D\) be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of \(D\) as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals.

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