Problem 49 - Single Variable Calculus

Question

Verify that the lateral surface area of a right circular cone of height \(h\) and base radius \(r\) is \(S=\pi r \sqrt{r^{2}+h^{2}}\) by evaluating a definite integral. Hint: The cone is generated by revolving the region bounded by \(y=(h / r) x, y=h\), and \(x=0\) about the \(y\) -axis.

Step-by-Step Solution

Answer

We verified the given formula for the lateral surface area of a right circular cone, \(S=\pi r \sqrt{r^{2}+h^{2}}\), by setting up a definite integral for the surface area of revolution and evaluating it. The function that forms the slant height of the cone is given by \(y=\frac{h}{r}x\), and we revolved it around the \(y\)-axis. After finding the inverse function and its derivative, we set up the definite integral, solved it, and simplified the result to obtain the given formula.

Step 1: Write down the equation for the lateral surface area of the cone

Based on the hint, the region that generates the cone is bounded by the following lines: 1. The line \(y = \frac{h}{r}x\) 2. The line \(y = h\) 3. The line \(x = 0\) The line \(y = \frac{h}{r}x\) will be the slant height of the cone when revolved around the \(y\)-axis.

Step 2: Set up the integral for surface area of revolution

When revolving a curve around an axis to form a surface of revolution, we can use the formula: \(A = 2 \pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx\) In our case, we are revolving the curve \(y=\frac{h}{r}x\) around the \(y\)-axis. So, we first need to express the function as the inverse \(x = f(y)\) and find its derivative. \(f(y) = x = \frac{r}{h}y\) Compute the derivative: \(f'(y) = \frac{r}{h}\) Now, we can set up the definite integral for the surface area: \(A = 2 \pi \int_{0}^{h} \frac{r}{h}y \sqrt{1 + \left(\frac{r}{h}\right)^2} dy\)

Step 3: Solve the integral

We will now solve the definite integral by removing all the constants and doing some simplification: \(A = 2 \pi \frac{r}{h} \int_{0}^{h} y\sqrt{1+\frac{r^2}{h^2}} dy\) \(A = 2 \pi r \sqrt{\frac{r^2}{h^2}+1} \int_{0}^{h} y dy\) Now, we find the antiderivative of \(y\) and evaluate it at the endpoints: \(A = 2 \pi r \sqrt{\frac{r^2}{h^2}+1} \left[\frac{y^2}{2}\right]_0^h\) \(A = 2 \pi r \sqrt{\frac{r^2}{h^2}+1}\left(\frac{h^2}{2}\right)\)

Step 4: Simplify and check the formula

We can simplify the result obtained from the definite integral to see if it matches the given formula for the lateral surface area of a cone, \(S=\pi r \sqrt{r^{2}+h^{2}}\): \(A = 2\pi r \left(\frac{h^2}{2}\right) \sqrt{\frac{r^2}{h^2}+1}\) \(A = \pi r h \sqrt{1+r^2/h^2}\) \(A = \pi r \sqrt{h^2+r^2}\) This matches the given formula, so we have successfully verified the lateral surface area of a right circular cone using a definite integral.

Key Concepts

Definite Integral

A definite integral is a mathematical tool used to compute the accumulation of quantities. It represents the area under a curve between two points on a graph. Physically, it can calculate quantities such as distance, area, and in some cases, surface areas of 3-dimensional objects formed by revolution. To find a definite integral, we first identify the function to integrate and the limits that define our interval.

In the context of our cone exercise, we used a definite integral to calculate the lateral surface area of the cone. We took advantage of the fact that the integral can handle the complex mathematics of curvature and accumulation in terms of simplifying a geometric shape like a cone. The limits of integration were from 0 to the height of the cone, and the function was set up to revolve around the axis, facilitating the measurement of area with precision.

Surface Area of Revolution

The surface area of revolution is an advanced concept in calculus. It refers to the surface area formed when a curve is revolved around a particular axis. To calculate it, we use a specific integral formula:

\[A = 2 \pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx\]

This formula takes in the original function defining the curve and its derivative. The incorporation of the derivative helps measure how the curve tilts, which is essential when determining how it spins into a surface. The multiplication by \(2\pi\) accounts for the circular path formed when revolving around an axis.

• Function of Integration: In our example, we used the line \(y = \frac{h}{r}x\) as the function to revolve, leading to the cone's surface.
• Revolution Axis: We revolved the curve around the y-axis, generating the cone's shape.
• Resultant Area: After evaluating the integral, we obtain the formula for the cone's lateral surface area.

Understanding this concept shows how calculus extends beyond simple area calculations to capture complex, real-world shapes.

Right Circular Cone

A right circular cone is a 3-dimensional geometric figure with a circular base and a single vertex aligned perpendicularly to the base. The `right` in "right circular cone" indicates that the apex (tip) of the cone is directly above the center of its base when viewed from the side.

• Components: It consists of a circular base, a vertex not on the base, and a lateral surface connecting the base to the vertex.
• Height and Radius: The distance from the base to the vertex in a straight vertical line is the height \(h\), while the distance from the center of the base to any point on the circumference is the radius \(r\).
• Slant Height: The slant height of the cone is the distance along the lateral surface from the base to the vertex. It's crucial for calculating the lateral surface area.

The lateral surface area of a right circular cone can be verified via calculus by revolving a line around an axis. The formula we verified, \(S = \pi r \sqrt{r^2 + h^2}\), encapsulates these dimensions into a concise expression that relates directly to both geometry and integral calculus.