Problem 79
Question
General volume formulas Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Frustum of a cone Find the volume of a truncated solid cone of height \(h\) whose ends have radii \(r\) and \(R\).
Step-by-Step Solution
Verified Answer
Question: Find the volume of a frustum of a cone with given height, \(h\), and radii \(r\) and \(R\) using integration.
Answer: The volume of the frustum of the cone is given by
$$V = \pi \left(-r^2 + 3Rr - 2R^2\right).$$
1Step 1: Set up the coordinate system
Choose cylindrical coordinates \((\rho, \phi, z)\), where \(\rho \geq 0, 0 \leq \phi \leq 2\pi\), and \(0 \leq z \leq h\). In this coordinate system, the equation of the surface can be obtained by combining the equations of the two ends of the frustum.
2Step 2: Find the equation of the bounding surface
In cylindrical coordinates, an equation of the lower circular end of the frustum is given by \(\rho = r\) when \(z = 0\). Similarly, an equation of the upper circular end of the frustum is given by \(\rho = R\) when \(z = h\). The lateral surface is defined as the line segment connecting the two ends.
Notice that the lateral surface runs diagonally in the \(z\) direction. If we start from the lower circular end with \(\rho = r\) at \(z=0\), we can see that the radius increases linearly with \(z\), reaching \(\rho = R\) at \(z=h\).
We can form a right triangle with base length \(z\), height \(\rho - r\), and hypotenuse \(h\). Hence, we have
$$\frac{\rho - r}{z} = \frac{R - r}{h} \Rightarrow \rho = r + \frac{z(R - r)}{h}.$$
This equation represents the lateral surface of the frustum.
3Step 3: Set up the triple integral
Now we'll set up the triple integral over the given region in cylindrical coordinates. Recall that the volume element in cylindrical coordinates is given by \(dV = \rho d\rho d\phi dz\). To find the limits of integration, first notice that \(0 \leq \phi \leq 2\pi\). For a given \(\phi\) and \(z\), \(\rho\) varies from the lower circular end at \(\rho = r\) to the lateral surface at \(\rho = r + \frac{z(R - r)}{h}\). Finally, \(z\) varies from the lower circular end at \(z = 0\) to the upper circular end at \(z = h\). Thus,
$$V = \int_{0}^{2\pi} \int_{0}^{h} \int_{r}^{r + \frac{z(R - r)}{h}} \rho \, d\rho \, dz \, d\phi.$$
4Step 4: Evaluate the triple integral
Evaluate the integral as follows:
$$V = \int_{0}^{2\pi} \int_{0}^{h} \left[ \frac{\rho^2}{2} \right]_{r}^{r + \frac{z(R - r)}{h}} dz \, d\phi = \int_{0}^{2\pi} \int_{0}^{h} \frac{1}{2}\left[(r + \frac{z(R - r)}{h})^2 - r^2\right] dz \, d\phi.$$
Now, integrate with respect to \(z\):
$$V = \int_{0}^{2\pi} \left[\frac{1}{2} \left(\frac{z^2(R-r)^2}{h^2} + 2rz(R-r)\right) - \frac{zr^2}{h}\right]_0^h d\phi = \int_{0}^{2\pi} \left[\frac{1}{2} \left(-r^2 + 3Rr - 2R^2\right)\right] d\phi.$$
Finally, integrate with respect to \(\phi\):
$$V = \left[\frac{1}{2} \left(-r^2 + 3Rr - 2R^2\right) \phi \right]_{0}^{2\pi} = \pi \left(-r^2 + 3Rr - 2R^2\right).$$
Thus, the volume of the frustum of the cone is
$$V = \pi \left(-r^2 + 3Rr - 2R^2\right).$$
Key Concepts
Cylindrical CoordinatesTriple IntegralFrustum of a Cone
Cylindrical Coordinates
Cylindrical coordinates are a powerful extension of polar coordinates used in three-dimensional space. In this system, a point in space is specified by three values: the radial distance \( \rho \), the angle \( \phi \), and the height \( z \). The radial distance \( \rho \) measures how far the point is from an origin line, similar to the radius in a circle. The angle \( \phi \) indicates the counterclockwise rotation from the positive x-axis, much like polar coordinates. The height \( z \) is the elevation of the point from a reference plane, typically the xy-plane.
This makes the cylindrical coordinate system especially convenient for solids with circular symmetry, such as cylinders and cones. In the given solution, cylindrical coordinates help simplify complex shapes like the frustum by adapting to its circular boundaries. The choice of cylindrical coordinates allows for easier calculation and integration when determining the volume of such shapes.
- \( \rho \geq 0 \)
- \(0 \leq \phi \leq 2\pi\)
- \(z \) is unbounded
This makes the cylindrical coordinate system especially convenient for solids with circular symmetry, such as cylinders and cones. In the given solution, cylindrical coordinates help simplify complex shapes like the frustum by adapting to its circular boundaries. The choice of cylindrical coordinates allows for easier calculation and integration when determining the volume of such shapes.
Triple Integral
A triple integral, as the name suggests, is an integral extended over three variables. It's an essential tool for finding volumes of three-dimensional regions. In cylindrical coordinates, the volume element is represented by \( dV = \rho \, d\rho \, d\phi \, dz \). This expression takes into account the radial nature of the system, with \( \rho \) ensuring the integration covers circular cross-sections effectively.
To set up a triple integral in cylindrical coordinates, you need to:
Each integration reflects the accumulation of infinitely small volumes to obtain the total volume. In our exercise, the integration order effectively captures the variation in the radius as we move along the height of the frustum.
To set up a triple integral in cylindrical coordinates, you need to:
- Establish limits for each of \( \rho \), \( \phi \), and \( z \).
- Integrate first over \( \rho \) (from the inner radius to the outer boundary defined by the problem).
- Then over \( z \) (from the base to the top of the object).
- Finally, over \( \phi \) (typically completing the full circle from 0 to \(2\pi\)).
Each integration reflects the accumulation of infinitely small volumes to obtain the total volume. In our exercise, the integration order effectively captures the variation in the radius as we move along the height of the frustum.
Frustum of a Cone
A frustum of a cone is a portion of a cone with its top part removed by slicing parallel to its base. It resembles a "truncated cone". The defining parameters are the radii of its circular ends and its height. In our problem, the base radius is \( r \), the top radius is \( R \), and the vertical height is \( h \). This shape maintains circular symmetry, making cylindrical coordinates ideal for analysis.
A frustum's volume can be found by integrating over this variable radius. In set up, the linear relationship between radius and height emerges from the truncated nature of the frustum: \[ \rho = r + \frac{z(R - r)}{h} \] This equation indicates that \( \rho \) (radius at any height \( z \)) changes linearly from \( r \) to \( R \).
Understanding the geometric nature of the frustum is key:
Analyzing a frustum with calculated equations helps form the right coordinate transformations, making solving the integral possible and accurate. Thus, an accurate method to quantify its volume.
A frustum's volume can be found by integrating over this variable radius. In set up, the linear relationship between radius and height emerges from the truncated nature of the frustum: \[ \rho = r + \frac{z(R - r)}{h} \] This equation indicates that \( \rho \) (radius at any height \( z \)) changes linearly from \( r \) to \( R \).
Understanding the geometric nature of the frustum is key:
- The lower base is at \( z = 0 \) with radius \( r \)
- The upper base is at \( z = h \) with radius \( R \)
- The sloped side ties these two ends, adjusting the radius according to the height \( z \)
Analyzing a frustum with calculated equations helps form the right coordinate transformations, making solving the integral possible and accurate. Thus, an accurate method to quantify its volume.
Other exercises in this chapter
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