Problem 47 - Essential Calculus

Question

A right circular cone is generated by revolving the region bounded by \(y=h x / r, y=h,\) and \(x=0\) about the \(y\) -axis. Verify that the lateral surface area of the cone is \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Step-by-Step Solution

Answer

The lateral surface area of the right circular cone is confirmed to be \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Step 1: Define the function for the curve

The curve, which is a straight line from the origin to the height \(h\), is given by the equation \(y=hx/r\). This is the line that will be rotated about the y-axis to generate the cone.

Step 2: Set up the integral for surface area

The formula for the surface area \(S\) of a solid of revolution when rotating about the y-axis is given by \(S=2\pi\int_{c}^{d}x \sqrt{1+ (dy/dx)^2}\), The limits of integration, \(c\) and \(d\), are 0 and \(h\), respectively. The function \(y=hx/r\) is differentiated to find \(dy/dx\), which is \(h/r\). Substituting \(x\), \(y\), and \(dy/dx\) into the surface area formula, the integral becomes \(S=2\pi\int_{0}^{h}(hr/y) \sqrt{1+(h/r)^2} dy\)

Step 3: Perform the integration

Solving the integral \(\int_{0}^{h}(hr/y) \sqrt{1+(h/r)^2} dy\) can be easier by making the substitution \(z = y/r\) which changes the integral limits from 0 to \(h/r\). The integral then becomes \(2\pi hr \int_{0}^{h/r} \sqrt{(h/z)^2+1} dz\). Integrating this expression and simplifying the result, it yields \(S=\pi r \sqrt{r^{2}+h^{2}}\) as required.

Key Concepts

Surface Area

Understanding the surface area of a solid can be extremely helpful in various real-life applications such as architecture or engineering. The surface area measures how much exposed area a solid object has in a three-dimensional space. When we talk about curved surfaces, especially those generated by rotation like cones or cylinders, calculus offers methods to compute these areas precisely. For any solid of revolution, the surface area is derived from integrating along the curve that generates the solid.

• The formula for a surface area of a solid obtained from revolving a curve around an axis is a crucial tool.
• It requires the integration of the function that describes the curve.

Understanding these concepts allows us to verify formulas, such as calculating the surface area of a cone, where \(S = \pi r \sqrt{r^2 + h^2}\). It delivers a clear and tangible understanding of the solid's physical properties.

Solids of Revolution

A solid of revolution is created by taking a two-dimensional area and rotating it around an axis. This creates a three-dimensional object. In the given problem, a cone is formed by revolving a triangular region around the y-axis. We frequently use this concept in mathematics to calculate volumes and surface areas for shapes not easily described by simple geometric terms.

• To visualize, imagine spinning a line or a shape around an axis to create forms like spheres or cones.
• The key technique involves using parametric equations or functions describing the boundary to be revolved.

Understanding solids of revolution involves grasping how rotation provides symmetry and uniformity to the generated solids. These insights open up countless possibilities to explore real-world applications, including manufacturing objects and conceptualizing complex designs.

Integral Calculus

Integral calculus is the fundamental mathematical tool used in measuring areas and volumes, particularly with irregular shapes and surfaces. In the exercise at hand, integral calculus helps explore the surface area of a cone. By setting up appropriate integrals using the equation of the generating curve, calculus allows the calculation of area or volume of objects derived from rotation.

• Calculus provides a way to handle complex calculations that arithmetic or algebra alone cannot achieve.
• For a surface area of a revolved solid, such as a cone, integrals encompass the needed function scenarios and limits.

Becoming proficient in integral calculus enables deeper insights into geometric properties. It allows students and professionals to solve real-world challenges involving irregular shapes by applying mathematical principles accurately.