Problem 56 - Calculus of a Single Variable

Question

Lateral Surface Area of a Cone 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 given statement is verified, the lateral surface area of the cone is $S=\pi r \sqrt{r^{2}+h^{2}}$

Step 1 - Definition of a right circular cone

A right circular cone is defined as a solid that forms when a right-angled triangle is rotated about one of its sides. In this case, the region bounded by \(y=h x / r,\) \(y=h,\) and \(x=0\) is being revolved around the \(y\)-axis.

Step 2 - Surface area of a solid of revolution by calculus

The surface area S of a solid of revolution is given by \(S = 2 \pi \int_{a}^{b} y ds\) where ds is the differential arc length, y is the radius of revolution which varies with x in this case and \(ds=\sqrt{1+(y')^{2}} dx\) where \(y'\) is the derivative of y with respect to x.

Step 3 - Application of formula

Applying the formula to our exercise, we have \(y = h x / r\) and \(y' = h / r\). Thus, \(ds = \sqrt{1+(h/r)^2} dx = r \sqrt{1+(x/r)^2} dx\), and \(S = 2 \pi \int_{a}^{b} y ds = 2 \pi \int_{0}^{r} (h x / r) \cdot r \sqrt{1+(x/r)^2} dx\). However the upper bound needs to be corrected, the upper bound would be \(h\), the height of the cone instead of \(r\), the radius of the base. This is because \(x\) was originally representing \(y\), so the bounds should be evaluated along the y-axis. So, \(S = 2 \pi \int_{0}^{h} (h x / r) \cdot r \sqrt{1+(x/r)^2} dx\).

Step 4 - Simplification and Final Result

Simplifying and solving, the integral yields the expression for the lateral surface area of a right circular cone which is \(S = \pi r \sqrt{r^{2}+h^{2}}\).

Key Concepts

Calculus

In the realm of mathematics, calculus is a field that specializes in studying continuous change. It's divided into two branches: differential calculus, which deals with rates of change and slopes of curves, and integral calculus, which deals with the accumulation of quantities and the areas under and between curves. Both branches come together to solve problems involving motion, area, volume, and many other concepts in physics and engineering. In the case of finding the lateral surface area of a cone, integral calculus comes into play as it allows us to calculate the total area by integrating over a curve. Integral calculus provides a systematic way to find areas, volumes, and other concepts that are not easily approachable with simple geometry.

Solid of Revolution

When we talk about a solid of revolution, we're referring to a three-dimensional object created by rotating a two-dimensional shape around an axis. The shapes that are revolved can be simple, like rectangles, or more complex, like curves from functions. The right circular cone in our exercise is an example of a solid of revolution, formed by rotating a right-angled triangle around one of its sides parallel to the y-axis. This method of creating solids is commonly reviewed in calculus because it gives rise to problems that can be solved using integration, allowing us to find volumes and surface areas of intricate shapes that would be otherwise difficult to calculate.

Surface Area Integration

Surface area integration is part of integral calculus and it is used to compute the surface area of solid objects. The fundamental idea is to break down the surface into infinitesimally small pieces, calculate the area of each piece, and then sum all these tiny areas together to get the total surface area. For a solid of revolution, this involves using a specific integral formula which accounts for the curved surface produced by rotation. The resulting integral takes the form \( S = 2 \pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx \), where \(f(x)\) is the function being revolved, and \(f'(x)\) is its derivative. The introduction of \(\pi\) reflects the circular nature of the surfaces being calculated. Calculating the lateral surface area of a cone is a practical example of surface area integration, as it involves evaluating an integral over the slant height of the cone.

Right Circular Cone

A right circular cone is a geometrical figure with a circular base and a smooth lateral surface that connects the base to a single point called the apex or vertex. The cone is termed 'right circular' because the line segment joining the apex to the center of the base is perpendicular to the base, and the base is a circle. In our context, the lateral surface area is of interest, which is the area of the cone's surface excluding the base. The formula for calculating the lateral surface area \(S\) is derived using calculus, specifically through surface area integration, as the cone is a solid of revolution created by rotating a right-angled triangle around one of the legs that's considered the axis of revolution. Understanding the geometrical and analytical properties of the right circular cone helps in solving problems involving this shape, such as finding its surface area and volume.