Question

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. Find the volume of a truncated solid cone of height \(h\) whose ends have radii \(r\) and \(R\).

Short answer

Answer: The volume of the truncated solid cone is given by \(V = \frac{(h_2 - h_1)R^2(h - h_2)^2 \pi}{2h_2^2}\).

Step-by-step solution

Set up the coordinate system and describe the geometry of the problem.

We will be using cylindrical coordinates, where \((r, \theta, z)\) represent radial distance, polar angle, and height, respectively. We are given a truncated solid cone, which consists of a smaller cone (with radius \(r\) and height \(h_1\)) removed from a larger cone (with radius \(R\) and height \(h_2\)). Both cones have the same apex, at a height \(h\) above the base of the larger cone. We need to find the volume of the region between the two cones.

Find the equations for bounding surfaces and limits of integration

Consider a cross-sectional slice of the two cones along the plane \(z=0\). We will find the following bounds for the triple integral: - For \(r\), it will range from 0 to the maximum radial distance at the cross-section, which is the line segment connecting the points where the smaller cone meets the larger cone. At the intersection of the large and small cones, we have \(\frac{R}{h_2} = \frac{r}{h-h_2}\), or \(r = \frac{R(h-h_2)}{h_2}\). So \(r\) will vary from 0 to \(\frac{R(h-h_2)}{h_2}\). - For \(\theta\), it will range from 0 to \(2\pi\). - For \(z\), it will range from the height of the smaller cone (\(h_1\)) to the height of the larger cone (\(h_2\)).

Set up the triple integral with cylindrical coordinates

The volume element in cylindrical coordinates is given by \(dV = r dz d\theta dr\). Our goal is to compute the volume of the region between the two cones, so the triple integral will be set up as: \(V = \int_{0}^{2\pi} \int_{0}^{\frac{R(h-h_2)}{h_2}} \int_{h_1}^{h_2} r dz d\theta dr\)

Evaluate the triple integral

Now we evaluate the integral: \(V = \int_{0}^{2\pi} \int_{0}^{\frac{R(h-h_2)}{h_2}} \left[\frac{r(z)|_{h_1}^{h_2}}{2}\right] d\theta dr = \int_{0}^{2\pi} \int_{0}^{\frac{R(h-h_2)}{h_2}} \frac{r(h_2 - h_1)}{2} dr d\theta\) Now we can integrate with respect to \(r\): \(V = \int_{0}^{2\pi} \left[\frac{(h_2 - h_1)r^2}{4}\right|_{0}^{\frac{R(h-h_2)}{h_2}} d\theta = \int_{0}^{2\pi}\frac{(h_2 - h_1)R^2(h - h_2)^2}{4h_2^2}d\theta\) Finally, integrating with respect to \(\theta\), we get: \(V = \left[\frac{(h_2 - h_1)R^2(h - h_2)^2}{4h_2^2}\theta\right|_{0}^{2\pi} = \frac{(h_2 - h_1)R^2(h - h_2)^2 \pi}{2h_2^2}\) So the volume of the truncated solid cone is given by: \(V = \frac{(h_2 - h_1)R^2(h - h_2)^2 \pi}{2h_2^2}\)

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are an extension of polar coordinates to three-dimensional space. They are especially useful for problems involving symmetries about an axis, making them ideal for solving problems related to circular or cylindrical objects.

In cylindrical coordinates, a point in space is represented by three values:

• Radial Distance ( ): Distance from the point to the z-axis.
• Polar Angle ( heta): Angle between the radial line and the positive x-axis, in the xy-plane.
• Height (z): Distance along the z-axis.

These coordinates simplify many integration problems where the symmetry of the involved geometries aligns well with circular profiles or rotational structures.

For example, when dealing with a truncated cone, as in our problem, cylindrical coordinates allow us to easily set up the necessary integrals by exploiting the symmetry around the central axis of the cone.

Triple Integral

Triple integrals extend the concept of integration to three dimensions. They are used to calculate volumes of three-dimensional objects or other properties, like mass, over a volume.

A triple integral in \(xyz\) coordinates is written as \(\int \int \int f(x, y, z) \, dx \, dy \, dz\).This integral is analogous to a double integral, which is used for areas, but instead, it is applied over a region in three-dimensional space. When dealing with cylindrical coordinates, the volume element \(dr\dtheta\d z,\)lies naturally within such coordinate systems. The function we integrate is often very simple, such as "1" when computing volume.

In our exercise concerning the truncated cone, the triple integral helps aggregate the small cylindrical sections of the cone into a total volume by summing up all the infinitesimal contributions from each volume element \(dr\dtheta\d z\). Finding the bounds for each variable is crucial, as these determine the specific region over which you are integrating.

Truncated Cone Volume

A truncated cone, also known as a frustum, is a cone with the top cut off parallel to the base. The volume of such a shape can be calculated using integration, making use of the solid's geometrical properties.

For a truncated cone with height \(h\),and varying radii at the ends (smaller radius \(r\)and larger radius\(R\)), the task is to find the volume between the top and bottom surfaces. In the given example, both ends are aligned perpendicularly to the central axis.

The volume is calculated by setting up a triple integral over the trapezoidal cross-section of the cone in cylindrical coordinates. The radial component \(r\)is carefully chosen based on the height because this varies linearly from \(r\)on the smaller base to\(R\)on the larger base. This allows the setup of bounds for \(r\)as a function of\(z\).Triple integration is then used to compile the incremental volumes from overlapping circular increments sliced along the cone's height.

Calculus Applications

Calculus plays a fundamental role in determining volumes of complex shapes through integration. This exercise demonstrated how calculus can handle real-life problems involving three-dimensional objects, like a truncated cone.

Applications of such calculus operations are seen across different fields, from physics to engineering. For example:

• Physics: Calculating the moment of inertia for a rotational object, which is essential in dynamic systems.
• Architecture: Designing objects with materials constraints, where precise volume calculations are necessary.
• Medical Imaging: Calculating the volume of organs from scanned data to aid in diagnostics.

The process of setting up and solving a triple integral helps in visualizing how different parts of a volume contribute to a total, and allows us to harness this understanding in practical applications. By choosing appropriate coordinates, such as cylindrical coordinates for a cone, the complexity of integration can be significantly reduced, highlighting the power and flexibility of calculus.