Problem 37 - Basic Technical Mathematics with Calculus

Question

The area of a surface of revolution from \(x=a\) to \(x=b\) is $$S=2 \pi \int_{a}^{b} y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x$$ Find the formula for the lateral surface area of a right circular cone of radius \(r\) and height \(h\).

Step-by-Step Solution

Answer

The lateral surface area of a cone is \(\pi r \sqrt{h^2 + r^2}\).

Step 1: Understand the Surface of Revolution Concept

A surface of revolution is generated by rotating a curve around an axis. In the case of a right circular cone, the line segment from the apex to the base's edge is revolved around the central axis to form the cone's lateral surface.

Step 2: Define the Function and Its Derivative

Consider a linear function from the apex (0,0) to the base (h,r), which is \(y = \frac{r}{h}x\). The derivative of this function with respect to \(x\) is \(\frac{dy}{dx} = \frac{r}{h}\).

Step 3: Substitute into the Surface Area Formula

Substitute the function \(y = \frac{r}{h}x\) and its derivative \(\frac{dy}{dx} = \frac{r}{h}\) into the given formula for surface area: \[ S = 2\pi \int_{0}^{h} \frac{r}{h}x \sqrt{1 + \left(\frac{r}{h}\right)^2} \, dx \]

Step 4: Simplify the Integral Expression

Simplify the expression inside the square root: \(\sqrt{1+(\frac{r}{h})^2} = \sqrt{\frac{h^2+r^2}{h^2}} = \frac{\sqrt{h^2+r^2}}{h}\). Substitute back into the integral, \[ S = 2\pi \int_{0}^{h} \frac{r}{h}x \cdot \frac{\sqrt{h^2+r^2}}{h} \, dx \] Which simplifies to: \[ S = \frac{2\pi r \sqrt{h^2+r^2}}{h^2} \int_{0}^{h} x \, dx \]

Step 5: Evaluate the Integral

To evaluate the integral, compute: \[ \int_{0}^{h} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{h} = \frac{h^2}{2} \] Therefore, the surface area \(S\) becomes: \[ S = \frac{2\pi r \sqrt{h^2+r^2}}{h^2} \times \frac{h^2}{2} = \pi r \sqrt{h^2+r^2} \]

Step 6: Finalize the Formula

The final formula for the lateral surface area of a right circular cone of radius \(r\) and height \(h\) is: \[ S = \pi r \sqrt{h^2+r^2} \]

Key Concepts

lateral surface area

The lateral surface area of a geometric shape refers to the area of all the surfaces of the shape excluding the areas of its bases. In the case of a right circular cone, this is the area of the curved surface that extends from the base to the apex.
To calculate the lateral surface area, one can imagine slicing the cone along a vertical line and unfolding it into a sector of a circle. The length of the arc of this sector equals the circumference of the cone's base, which is \(2 \pi r\), and the radius of the sector is the slant height of the cone, \(\sqrt{h^2 + r^2}\).
Thus, the lateral surface area \(S\) is given by the formula:
\[ S = \pi r \sqrt{h^2 + r^2} \]

• \(\pi\) is the constant approximately equal to 3.14159.
• \(r\) is the radius of the base of the cone.
• \(h\) is the height of the cone.
• \(\sqrt{h^2 + r^2}\) is the slant height of the cone.

The lateral surface area doesn't include the base area, but only the sloped portion wrapping around from the base to the apex.

right circular cone

A right circular cone is a three-dimensional geometric figure with a circular base and a pointed apex or vertex perpendicular to the center of the base.
This specific cone type features a symmetry axis, referred to as the axis of the cone, which is aligned perpendicularly to its base. This configuration grants the cone its name since all cross-sections parallel to the base will remain circular.
The right circular cone has two prime dimensions:

• The radius \(r\), which describes the distance from the center to the perimeter of the circular base.
• The height \(h\), defined as the perpendicular distance from the base to the apex or vertex.

Additionally, the slant height, \(\sqrt{h^2 + r^2}\), connects the apex to any given point on the edge of the base, and it forms part of the triangle comprising the height and the radius in a right-angled triangle setup. This unique height configuration is what differentiates right circular cones from oblique cones, which have an apex that is offset from the base's center.

integral calculus

Integral calculus is a branch of mathematical analysis focused on integration and its myriad of applications. Integration is essentially the inverse operation of differentiation, where the aim is to compute an antiderivative or integral.
In the context of surfaces of revolution, integral calculus allows us to determine the area of these surfaces by summing infinitesimally small surface elements revolved around an axis.
This process involves:

• The function \(y = \frac{r}{h}x\), representing the linear path from the cone's apex to its base.
• The derivative \(\frac{dy}{dx} = \frac{r}{h}\) needed for the integration formula.

By setting up and evaluating the integral
\[ S = 2\pi \int_{a}^{b} y \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} \, dx \]
we can deduce the formula for the lateral surface area of the cone.
Integral calculus thus provides a comprehensive tool for deriving area measurements that would otherwise be challenging to calculate through simple geometry alone. It summarizes many small elements into a singular result, exemplifying its powerful utility in mathematics.