Problem 77 - Calculus Early Transcendentals

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. Cone Find the volume of a solid right circular cone with height \(h\) and base radius \(r\).

Step-by-Step Solution

Answer

Question: Find the volume of a solid right circular cone with height \(h\) and base radius \(r\) using integration. Answer: The volume of the solid right circular cone with height \(h\) and base radius \(r\) is \(V = \frac{\pi h r^2}{3}\).

Step 1: Set up the coordinate system and equation of bounding surfaces

In the cylindrical coordinate system, the equation of the base can be represented as \(z = 0\). The equation of the side surface can be obtained by considering a right triangle formed by the height (\(h\)), base radius (\(r\)), and a slant height. The slope of this triangle can be expressed as \(\frac{h}{r}\). So, the equation of the side is given by \(z = \frac{h}{r} \cdot r = h - \frac{h}{r} \cdot r\). Now that we have the equations of the bounding surfaces, we can set up a triple integral.

Step 2: Set up the triple integral

Using the equations of the bounding surfaces, we can write the volume of the cone as a triple integral: \(V = \iiint\limits_D dV\). We need to evaluate this triple integral in the cylindrical coordinate system, which has a differential volume element \(dV = r \, dz \, dr \, d\theta\). Therefore, the integral becomes: \(V = \iiint\limits_D r \, dz \, dr \, d\theta\).

Step 3: Define the limits of integration

We now need to find the limits of integration. Since the cone is symmetric about the z-axis, we can integrate \(\theta\) from \(0\) to \(2\pi\): \(0 \leq \theta \leq 2\pi\). For the radius, \(r\) ranges from the center of the cone (\(r = 0\)) to the edge of the base (\(r = r\)): \(0 \leq r \leq r\). For the height, \(z\) varies from the apex (\(z = 0\)) to the side surface (\(z = h - \frac{h}{r} \cdot r\)): \(0 \leq z \leq h - \frac{h}{r} \cdot r\). With these limits, we can set up the triple integral: \(V = \int_{0}^{2\pi}\int_{0}^{r}\int_{0}^{h-\frac{h}{r}\cdot r} r \, dz \, dr \, d\theta\).

Step 4: Evaluate the triple integral

First, integrate with respect to \(z\): \(V = \int_{0}^{2\pi}\int_{0}^{r} \left[ \frac{r^2}{2}z \right]_{0}^{h-\frac{h}{r}\cdot r} \, dr \, d\theta = \int_{0}^{2\pi}\int_{0}^{r} \frac{r^2}{2}(h-\frac{h}{r}\cdot r) \, dr \, d\theta\). Now, integrate with respect to \(r\): \(V = \int_{0}^{2\pi} \left[\frac{r^3}{6}h - \frac{r^4}{8}h \right]_{0}^{r} \, d\theta = \int_{0}^{2\pi} \left(\frac{r^3}{6}h - \frac{r^4}{8}h \right) \, d\theta\). Finally, integrate with respect to \(\theta\): \(V = \left[ \frac{h(2\pi)(r^3)}{6} - \frac{h(2\pi)(r^4)}{8} \right]_{0}^{2\pi} = \frac{\pi h r^2}{3}\). So, the volume of the solid right circular cone with height \(h\) and base radius \(r\) is: \(V = \frac{\pi h r^2}{3}\).

Key Concepts

Triple Integral

A triple integral is like doubling down on integrals. Instead of just one or two layers, it adds another dimension. This is especially useful when figuring out volumes of 3D shapes like cones. By stacking tiny 'slices' of the shape, we combine them to find its full volume.
Imagine you are stacking pancakes in the shape of a cone. Each integral helps you decide:

• How wide each pancake slice is (the circular area).
• How high you go (the height of the cone).
• All around the cone (the rotation to complete the full shape).

This three-part process helps to precisely calculate volumes in complex shapes and is a handy tool in calculus, especially when dealing with solids bounded by curves in space.

Cylindrical Coordinates

Cylindrical coordinates are like a new set of directions. Instead of the traditional x, y, z coordinates, we use

• radial distance (\(r\)) from a fixed axis,
• angle (\(\theta\)) around the axis,
• height (\(z\)).

This system is perfect for shapes with circular bases, like cones. By using cylindrical coordinates, we can simplify the process of finding the volume even more.
Think of how a cone has a circular base and rises straight up. Using the radial distance and angle, you can define any point on the cone's surface. This makes volume calculations much simpler by aligning the integral with the natural geometry of the cone.

Cone Volume Formula

The cone volume formula is a classic geometry equation: \(V = \frac{1}{3}\pi r^2 h\). But how is it derived? Well, this formula comes from using integration to sum up an infinite number of infinitesimally thin disks stacked up to form the cone. Each disk has a small piece of the cone's base.

• Each disk's volume is like a tiny pancake.
• The area of the disk is \(\pi r^2\).
• Stack them up varying the height, all the way from the apex to the base.

Integration lets us sum all these tiny disk volumes, bringing us to the beautifully simple result of \(\frac{1}{3}\pi r^2 h\). This connects integrals and geometry, showing the mathematical elegance behind this standard formula.

Integration in Calculus

Integration is a fundamental concept in calculus. It reverses differentiation, finding the original amount if you know the rate of change. With integration, you can find areas below curves and volumes inside shapes.

• In calculus, integration provides a way to sum up small changes.
• This is especially useful in finding physical quantities like volumes.

Think of it as a way to gather all the tiny changes to see the whole picture.
In the case of a cone, integration helps express the volume by summing cylindrical slices of variable height. The power of calculus and integration allows this abstract summation to result in very real, calculable volumes. This shows not only the practicality of calculus but its elegance in describing the natural world.