Question

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Find the volume of a right circular cone with height \(h\) and base radius \(r\)

Short answer

The volume of a right circular cone with height \(h\) and base radius \(r\) is given by the formula \(\frac{1}{3} \pi r^3 h\).

Step-by-step solution

Convert to cylindrical coordinates

We first change to cylindrical coordinates, which are more suitable for this problem. In cylindrical coordinates, a point \((x, y, z)\) is represented by \((\rho, \phi, z)\), where \(\rho\) is the distance from the point to the \(z\)-axis, \(\phi\) is the angle in the \(xy\) plane measured from the positive \(x\)-axis, and \(z\) is the height above the \(xy\) plane.

Find equations for the bounding surfaces

To find the equation of the cone, we note that the slope of the cone is constant and equal to \(\frac{h}{r}\). In cylindrical coordinates, the equation of the cone is given by: \(z = \frac{h}{r} \rho\) The bounding surfaces are given by the following equations: 1. The cone: \(z = \frac{h}{r} \rho\) 2. The base: \(z = 0\) 3. The upper limit of radial distance: \(\rho = r\) 4. The angle \(\phi\) varies from \(0\) to \(2\pi\)

Set up the volume integral

In cylindrical coordinates, the volume integral becomes: \(V = \int\int\int \rho \, d\rho \, d\phi \, dz\) Now, we must set the limits of integration. We know: \(0 \le \rho \le r\) \(0 \le z \le \frac{h}{r} \rho\) \(0 \le \phi \le 2\pi\) So, the integral becomes: \(V = \int_{0}^{2\pi} \int_{0}^{r} \int_{0}^{\frac{h}{r}\rho} \rho \, dz \, d\rho \, d\phi\)

Evaluate the integral

We can now evaluate the integral for the volume: \(V = \int_{0}^{2\pi} \int_{0}^{r} \int_{0}^{\frac{h}{r}\rho} \rho \, dz \, d\rho \, d\phi\) \(V = \int_{0}^{2\pi} \int_{0}^{r} \left[\frac{h}{r} \rho^2\right]_0^{\frac{h}{r}\rho} \, d\rho \, d\phi\) \(V = \int_{0}^{2\pi} \int_{0}^{r} {\frac{h}{r} \rho^3} \, d\rho \, d\phi\) \(V = \int_{0}^{2\pi} \left[\frac{h}{4r} \rho^4\right]_0^{r} \, d\phi\) \(V = \int_{0}^{2\pi} {\frac{h}{4} r^3} \, d\phi\) \(V = \left[\frac{h}{4} r^3 \phi\right]_0^{2\pi}\) \(V = \frac{1}{3} \pi r^3 h\) The volume of a right circular cone with height \(h\) and base radius \(r\) is \(\frac{1}{3} \pi r^3 h\).

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are extremely useful when dealing with problems that have symmetry around an axis, such as a cone. Instead of working with Cartesian coordinates \((x, y, z)\), cylindrical coordinates \((\rho, \phi, z)\) simplify the process by introducing more appropriate parameters:

• \( \rho \): Represents the radial distance from the \(z\)-axis. Think of it as how far your point is from the vertical center line of the cone.
• \( \phi \): Is the angular coordinate. It's the angle around the \(z\)-axis, measured in the plane between the positive \(x\)-axis and our point.
• \( z \): Corresponds to the height or altitude of the point above the \(xy\) plane.

These coordinates are particularly convenient for cones because they match the natural symmetry and simplify the mathematics of volume integration.
Converting to cylindrical coordinates lets us efficiently describe the cone and set up the necessary calculations.

Volume Integral

A volume integral helps find the total volume of a region. For a cone, these integrals are simplified in cylindrical coordinates. To perform a volume integral:

• Set up the integral: The formula \( V = \int\int\int \rho \, d\rho \, d\phi \, dz \) accounts for the volume through tiny cylindrical sections.
• Think of small elements: The \( \rho \, d\rho \, d\phi \, dz \) term represents a tiny cylindrical slice of the cone. By summing up these slices over the defined bounds, we obtain the cone's total volume.

When solving the integral, you process it layer by layer. Start integrating with respect to \( z \), then \( \rho \), and finally \( \phi \). This structured approach makes it easier to solve complex volume problems using familiar calculus techniques.

Bounding Surfaces

Bounding surfaces define the limits of our integration and describe the region inside the cone:

• Cone surface: In cylindrical coordinates, the equation \( z = \frac{h}{r} \rho \) represents the slanted surface of the cone. The slope indicates how the height changes with the radius.
• Base plane: The bottom surface is simply \( z = 0 \), the base of the cone.
• Radial boundary: The cone's largest width is given by \( \rho = r \), setting the boundary for how far out \( \rho \) can go.
• Angular range: \( \phi \), the angle, travels through a complete circle from \( 0 \) to \( 2\pi \), covering the whole cone around its axis.

Understanding these surfaces is crucial as they determine where the volume exists within the space, ensuring the integration confines precisely to the cone's shape.

Integration Limits

The process of determining integration limits is key to correctly setting the volume integral. For a cone, these limits are:

• Radial limits: \( 0 \leq \rho \leq r \). This means the radial distance starts from the center and extends outwards to the cone's base radius.
• Vertical limits: \( 0 \leq z \leq \frac{h}{r} \rho \). This accounts for the slant of the cone, with \( z \) increasing with \( \rho \). The upper bound for \( z \) varies as we move radially outward, defined by the cone’s surface slope.
• Angular limits: \( 0 \leq \phi \leq 2\pi \). The integration rotates completely around to cover the circular cross-section at each height.

Understanding each limit ensures the integral only covers points inside the cone, providing the correct volume by considering the entirety of its spatial extent.