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Chapter 6: Problem 26

Testing the new definition The lateral (side) surface area of a cone of height \(h\) and base radius \(r\) should be \(\pi r \sqrt { r ^ { 2 } + h ^ { 2 } }\) , the semiperimeter of the base times the slant height. Show that this is still the case by finding the area of the surface generated by revolving the line segment \(y = ( r / h ) x , 0 \leq x \leq h ,\) about the \(x\) -axis.

Short Answer

Expert verified
The lateral surface area formula is correctly shown as \( \pi r \sqrt{r^2 + h^2} \).

Step by step solution

01

Understand the Geometry

We need to find the lateral surface area of a cone by revolving the line segment around the x-axis. The line segment is given by the equation \( y = \frac{r}{h} x \), over the interval \( 0 \leq x \leq h \). The segment rotates around the x-axis to form a cone with height \( h \) and base radius \( r \).
02

Set Up the Surface Area Integral

The lateral surface area of a solid formed by rotating a curve \( y = f(x) \) about the x-axis is given by the integral \( \int_{a}^{b} 2 \pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). Here, \( y = \frac{r}{h}x \) and \( \frac{dy}{dx} = \frac{r}{h} \).
03

Substitute and Simplify

Substituting \( y = \frac{r}{h}x \) and \( \frac{dy}{dx} = \frac{r}{h} \) into the integral gives us: \[ \int_{0}^{h} 2 \pi \left( \frac{r}{h}x \right) \sqrt{1 + \left( \frac{r}{h} \right)^2} \, dx \].Simplifying inside the integral, we have:\[ \int_{0}^{h} 2 \pi \frac{r}{h}x \sqrt{1 + \frac{r^2}{h^2}} \, dx = \int_{0}^{h} 2 \pi \frac{r}{h}x \sqrt{\frac{h^2 + r^2}{h^2}} \, dx \].
04

Factor and Simplify Further

Since \( \sqrt{\frac{h^2 + r^2}{h^2}} = \frac{\sqrt{h^2 + r^2}}{h} \), the integral becomes: \[ \int_{0}^{h} 2 \pi \frac{r}{h}x \cdot \frac{\sqrt{h^2 + r^2}}{h} \, dx = \int_{0}^{h} 2 \pi \frac{r \sqrt{h^2 + r^2}}{h^2} x \, dx \].Factor out constants from the integral:\[ = \frac{2 \pi r \sqrt{h^2 + r^2}}{h^2} \int_{0}^{h} x \, dx \].
05

Integrate and Evaluate

Evaluate the integral \( \int_{0}^{h} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{h} = \frac{h^2}{2} \). Substitute back: \[ \frac{2 \pi r \sqrt{h^2 + r^2}}{h^2} \cdot \frac{h^2}{2} = \pi r \sqrt{h^2 + r^2} \].
06

Conclusion

We have shown that the lateral surface area of the cone is indeed \( \pi r \sqrt{r^2 + h^2} \), which is the expression derived from the definition of the semiperimeter times the slant height.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus Integration
Calculus integration is a powerful mathematical tool used to calculate various areas and volumes, such as the lateral surface area of a cone. By integrating, we can sum infinitely small elements along curves and surfaces. In this exercise, we calculated the surface area through an integral that considers the line segment's revolution around an axis. The integral formula for surface area involves both the function and its derivative, simplifying it using calculus rules to find the final surface area of rotation.
Surface Area Calculation
Calculating the surface area for curved shapes, like a cone, is different from straightforward shapes. We take into account not just the base, but all sloping sides due to their curvature. The formula derived from integrating effectively sums up all those infinitesimal elements on the surface of the sphere. We use a specific integral form \(\int_{a}^{b} 2 \pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\) to handle such a calculation, accommodating for both area and curve.
  • This involves both the function itself and the derivative.
  • The limits of integration define the length of the segment revolved.
In simple terms, we rotate a line to form the cone and calculate every tiny segment to find the area.
Revolution About Axis
The process of generating a 3D shape, like a cone, from a 2D line by revolving it around an axis is a fundamental concept in geometry known as revolution about an axis. In this task, the line equation \(y = \frac{r}{h}x\) is rotated around the x-axis to create the cone. This method highlights how rotating simple geometry can result in complex volumes and surfaces. By understanding the axis of revolution, students can visualize the transformation from a flat line to a full three-dimensional solid.
Slant Height
The slant height of a cone is a key measurement that helps in calculating its lateral surface area. It represents the shortest distance from the apex of the cone to any point on the edge of the base. In the context of the formula \(\pi r \sqrt{r^2 + h^2} \), the term \(\sqrt{r^2 + h^2}\) represents the slant height. This takes into account both the height and radius of the cone, forming a right triangle where the slant height is the hypotenuse. Understanding slant height is crucial for grasping how the cone's surface area is derived geometrically.
Geometry of Solids
Understanding the geometry of solids like cones involves recognizing their shape, size, and properties. For a cone, key features include the base, height, radius, and slant height. These are used to derive important formulas, such as the lateral surface area or total volume. In this exercise, we focused on how the geometry changes during revolution and how the straight-line segment morphs into a cone. Mastering these spatial transformations is essential for solving complex geometry problems involving solids.

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Most popular questions from this chapter

Find the volume of the solid generated by revolving each region about the \(y\)-axis. The region enclosed by the triangle with vertices \((0,1),(1,0),\) and \((1,1)\)

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