Problem 65 - Calculus Early Transcendentals

Question

Verify that the volume of a right circular cone with a base radius of \(r\) and a height of \(h\) is \(\pi r^{2} h / 3 .\) Use the region bounded by the line \(y=r x / h,\) the \(x\) -axis, and the line \(x=h,\) where the region is rotated around the \(x\) -axis. Then (a) use the disk method and integrate with respect to \(x,\) and (b) use the shell method and integrate with respect to \(y\).

Step-by-Step Solution

Answer

Q: Verify the volume formula for a right circular cone with base radius \(r\) and height \(h\) using the disk and shell methods. A: By using the disk method, we found that the volume formula is \(\pi \frac{1}{3}r^2h\) and by using the shell method, we also arrived at the same volume formula \(\pi \frac{1}{3}r^2h\). Therefore, the volume formula for a right circular cone is verified to be \(\pi \frac{1}{3}r^2h\).

Step 1: Identify the region

The region under consideration lies between the line \(y=\frac{rx}{h},\) the \(x\)-axis and the line \(x=h\). When this region is rotated around the \(x\)-axis, it forms a right circular cone with base radius \(r\) and height \(h\).

Step 2: Disk Method Integration

Now, we will use the disk method while integrating with respect to \(x\). The disk method formula is \(V=\pi\int^{b}_{a} [f(x)]^2 dx,\) where \(V\) is the volume, and \(f(x)\) is the given function. The function for the given line is \(f(x) = \frac{rx}{h}\). In this case, for disk method, we have \(a = 0\) and \(b = h\). Plug these values and the function into the disk method formula: \(V = \pi\int^{h}_{0} [\frac{rx}{h}]^2 dx\)

Step 3: Calculate Disk Method Volume

Now, we will calculate the volume using the integral derived above: \(V = \pi\int^{h}_{0} [\frac{r^2 x^2}{h^2}] dx\) Evaluate the integral: \(V = \pi \frac{r^2}{h^2} \int^{h}_{0} x^2 dx\) \(V = \pi \frac{r^2}{h^2} [\frac{1}{3}x^3] |_0^h\) \(V = \pi \frac{r^2}{h^2} [\frac{1}{3}h^3]\) \(V = \pi \frac{1}{3}r^2h\), which is the volume formula for a cone. Now, we will use the shell method to verify the volume formula.

Step 4: Shell Method Integration

For the shell method, we will integrate with respect to \(y\). The shell method formula is \(V=2\pi\int^{B}_{A}yf(y) dy,\) where \(V\) is the volume, and \(f(y)\) is the expression of the variable \(x\) in terms of \(y\). The region boundaries for shell method are \(A = 0\) and \(B = r\). We need to find the expression of \(x\) in terms of \(y\) from the given function \(y = \frac{rx}{h}\): \(x = \frac{hy}{r}\) Plug these values and the expression into the shell method formula: \(V = 2\pi\int^{r}_{0} y\frac{hy}{r} dy\)

Step 5: Calculate Shell Method Volume

Now, we will calculate the volume using the integral derived above: \(V = 2\pi\int^{r}_{0} \frac{h^2 y^2}{r} dy\) Evaluate the integral: \(V = 2\pi\frac{h^2}{r} \int^{r}_{0} y^2 dy\) \(V = 2\pi\frac{h^2}{r} [\frac{1}{3}y^3] |_0^r\) \(V = 2\pi\frac{h^2}{r} [\frac{1}{3}r^3]\) \(V = \pi\frac{1}{3}r^2h\), which is the volume formula for a cone. Both the disk method and the shell method result in the same volume formula for a right circular cone, \(\pi \frac{1}{3}r^2h.\) Therefore, we have verified the volume formula as required.

Key Concepts

Disk Method

The disk method is a technique in integral calculus used to find the volume of a solid of revolution. Imagine slicing the solid into tiny disks (similar to very thin cylinders). Each disk is perpendicular to the axis of rotation. The volume of these disks is added together to approximate the total volume.

For a function rotated around the x-axis, the formula is \[V = \pi \int_{a}^{b} [f(x)]^2 \, dx.\] Here, \(f(x)\) represents the function being rotated, and \([a, b]\) is the range of integration.

• "\(\pi\)" accounts for the circular area of the disk.
• "\([f(x)]^2\)" is the radius squared, corresponding to the y-value of the function.

Applying it to verify the volume of a cone, the function \(y = \frac{rx}{h}\) is rotated from \(a = 0\) to \(b = h\). The evaluation gives the familiar \(\pi \frac{1}{3} r^2 h\). This demonstrates how tiny disks, when summed, reveal the cone's complete volume.

Shell Method

The shell method offers another approach to finding the volume of solids of revolution, especially useful when rotating around axes other than where the solid naturally lies. Instead of disks, this technique uses cylindrical shells.

The volume formula for the shell method is \[ V = 2\pi \int_{A}^{B} y f(y) \, dy,\] where \(f(y)\) is the distance function related to the rotation axis.

• "\(2\pi\)" represents the circumference of the cylindrical shell.
• "\(y\cdot f(y)\)" accounts for the cylindrical shell's height and radius.

For our cone's volume, the function \(x = \frac{hy}{r}\) forms cylindrical shells when integrated from \(A = 0\) to \(B = r\), verifying \(\pi \frac{1}{3} r^2 h\). This method helps visualize the shells building up to form the desired volume.

Integral Calculus

Integral calculus is essential for determining areas, volumes, and sums. It broadens the scope from calculating simple geometric shapes to more complex curves and solids.

By using integration, we can handle irregular shapes and forms that aren't easily broken down into conventional geometric objects. Integration takes into account:

• The sum of infinitely small quantities.
• The calculation of areas and volumes under curves and surfaces.

For the cone, integral calculus allows us to derive its volume formula \(\frac{1}{3} \pi r^2 h\) by handling its linear boundary \(y = \frac{rx}{h}\). This showcases the power of integral calculus to resolve complex shapes.

Rotational Volume

Rotational volume refers to the space occupied by a 3D object created by rotating a 2D shape about an axis. Understanding rotational volumes is key to engineering, design, and various sciences.

Methods like disk and shell allow us to compute these volumes by considering how the 2D region is swept into a 3D form:

• "Disk Method" uses circular cross-sections along the axis of rotation.
• "Shell Method" uses cylindrical shells with closed loops around the axis.

For our example, rotating the region defined by \(y = \frac{rx}{h}\) covers an area that solidifies into a cone. Both methods compute the cone's volume as \(\pi \frac{1}{3} r^2 h\). This exact volume calculation connects us to concepts crucial for calculating real-world shapes and structures.