Chapter 11: Problem 36
A cone The line segment joining the origin to the point \((h, r)\) is revolved about the \(x\) -axis to generate a cone of height \(h\) and base radius \(r .\) Find the cone's surface area with the parametric equations \(x=h t, y=r t, 0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi r(\) slant height\().\)
Short Answer
Step by step solution
Identify Parametric Equations
Express Derivatives in Terms of t
Calculate Differential Arc Length
Calculate Surface Area
Complete the Integration
Verify with Geometry Formula
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Consider the parametric equations \( x = ht \) and \( y = rt \), where \( 0 \leq t \leq 1 \). Here, \( t \) acts as the parameter that controls the movement from the origin \( (0,0) \) to the point \( (h,r) \).
- Each point on the line segment from the origin to \((h, r)\) is described as \( t \) varies between 0 and 1.
- These equations effectively translate the linear path into a manageable form for further calculations, such as finding the surface area of the cone, by focusing on how \( x \) and \( y \) change over the interval.
Differential Arc Length
Given the derivatives \( \frac{dx}{dt} = h \) and \( \frac{dy}{dt} = r \) from our parametric equations, we find:
\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \sqrt{h^2 + r^2} \, dt\]
- This formula captures how the curve stretches and bends between every infinitesimal increment of \( t \).
- By squaring and adding the derivatives, we ensure that \( ds \) accounts for changes in both directions (x and y) as we trace the curve.
- The constant \( \sqrt{h^2 + r^2} \) reflects that for this linear segment, the full length is uniform for any subdivided section, which simplifies its integration over \( t \).
Surface Area of Revolution
To calculate the surface area, we use the formula:
\[A = \int 2 \pi y \, ds\]Substitute \( y = rt \) and the simplified differential arc length \( ds = \sqrt{h^2 + r^2} \, dt \):
\[A = \int_{t=0}^{t=1} 2 \pi rt \sqrt{h^2 + r^2} \, dt = 2 \pi r \sqrt{h^2 + r^2} \int_{t=0}^{t=1} t \, dt\]
- The calculation reveals how stretching the infinitesimal arc length \( ds \) around the \( x \)-axis contributes to the surface area.
- The integration of \( t \) from 0 to 1 calculates the cumulative effect of rotating each infinitesimal element of the line, culminating in half the full area \( \left( \frac{1}{2} \right) \).
- By substituting values directly into the integral and performing the calculations, we correctly verify the result matches the geometric formula for a cone's lateral surface area \( \pi r \sqrt{h^2 + r^2} \).