Problem 52 - Single Variable Calculus

Question

Verify the formula for the volume of a right circular cone by finding the volume of the solid obtained by revolving the triangular region with vertices \((0,0),(0, r)\), and \((h, 0)\) about the \(x\) -axis.

Step-by-Step Solution

Answer

The solid formed by revolving the triangular region with vertices \((0,0)\), \((0,r)\), and \((h,0)\) about the \(x\)-axis is a right circular cone. To find its volume, we use the method of cylindrical shells; the volume of the cone is given by: \[ V = \int_0^h 2\pi R(x) \cdot dx \] After evaluating the integral and simplifying the expression, we obtain the volume formula for a right circular cone: \[ V = \pi r h^2 \]

Step 1: Set up the integral for the triangular region

Since we are revolving the triangular region around the \(x\)-axis, we will use the horizontal height, \(y\), for each infinitesimal cylinder. The height of the cylinder will be equal to the thickness along the \(x\)-axis, which is denoted as \(dx\). Let's first find the equation of the line defining the triangular region. The equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x-x_1) \] Using the points \((0,r)\) and \((h,0)\): \[ y - r = \frac{-r}{h}(x - 0) \] \[ y = -\frac{r}{h}x + r \]

Step 2: Determine the equation for the radius of each cylindrical shell

The radius, \(R\), of each cylindrical shell at a given \(x\) will be the height, \(y\), of the triangle at that point, which is given by the equation we derived above: \[ R(x) = -\frac{r}{h}x + r \]

Step 3: Apply the method of cylindrical shells

Now we have all the necessary elements to determine the volume using the method of cylindrical shells: \[ V = \int_0^h 2\pi R(x) \cdot dx \] Plug in the equation for \(R(x)\): \[ V = \int_0^h 2\pi \left(-\frac{r}{h}x + r\right) dx \]

Step 4: Evaluate the integral

Now we'll integrate the expression with respect to \(x\): \[ V = 2\pi \int_0^h \left(-\frac{r}{h}x + r\right) dx \] \[ V = 2\pi \left[-\frac{r}{2h}x^2 + rx \right]_0^h \] Evaluate the limits: \[ V = 2\pi \left[-\frac{r}{2h}h^2 + rh\right] - 0 \] \[ V = 2\pi \left[-\frac{1}{2}rh^2 + rh^2 \right] \]

Step 5: Simplify

Simplify the expression to find the final volume: \[ V = 2\pi \left[\frac{1}{2}rh^2\right] \] \[ V = \pi r h^2 \] So the volume of the solid formed by revolving the triangular region around the \(x\)-axis is \( V = \pi r h^2 \), which is the formula for the volume of a right circular cone.

Key Concepts

Cylindrical Shells

When we talk about calculating the volume of a cone using the method of cylindrical shells, think about peeling an orange. If you revolve a triangle around one of its axes, it creates circular layers. Each layer resembles a cylindrical shell. This approach is particularly useful for objects with circular symmetry around an axis.
To find the volume using cylindrical shells, we take many thin shell-like slices of the cone. These slices are essentially small cylinders that, when summed together, give us the volume of the solid. The key steps in using cylindrical shells are:

• Determine the function that represents the shape.
• Set up an integral that calculates all the shells' individual volumes.
• Use the shell radius and height to construct the integral.

In our example, the function from the triangle's line equation determines the shell's height, while the change in thickness, or the infinitesimal part, is treated using calculus.

Integral Calculus

Integral calculus is a fundamental tool for finding areas and volumes. It allows us to sum an infinite number of infinitesimally small parts to get a whole. The volume of a cone is a classical example, where integration combines countless thin cylinders into one volume.
In practice, to form the integral, we consider elements like width, height, and radius — each contributing to a solid's overall dimension.

• Firstly, if the solid is rotated around an axis, the equation of the region might determine the shape of each piece.
• Next, setting up the integral involves the product of function terms like length and width.
• Finally, evaluating this integral involves calculus rules — like substitution or part-by-part integration.

By doing the integral calculus for our exercise, we smoothly changed from dividing into many small parts to calculating a precise total measurement.

Parametric Revolutions

A parametric revolution involves revolving a planar shape around an axis to create a three-dimensional solid. In practical terms, imagine rotating a piece of paper to form a cone, like how this exercise spins the triangle.
To map out a parametric revolution:

• Define the boundary lines of the shape you are revolving.
• Decide which axis the shape revolves around — this influences the formula and calculations.
• Establish the integral boundaries, reflecting the shape's orientation and position.

In the example provided, the triangle revolves about the x-axis. Its vertices help specify the integral's limits (from 0 to the given height). The parametric approach enables visualization and a deeper understanding of generating cones or similar solids by rotation.