/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A cone frustum The line segment ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 37

A cone frustum The line segment joining the points \((0,1)\) and \((2,2)\) is revolved about the \(x\) -axis to generate a frustum of a cone. Find the surface area of the frustum using the parametrization \(x=2 t, y=t+1,0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi\left(r_{1}+r_{2}\right)(\text { slant height). }\)

Short Answer

Expert verified
The surface area of the frustum is \(3\pi\sqrt{5}\).

Step by step solution

01

Understand Parametrization

The given parametrization for the line segment is \( x = 2t \) and \( y = t + 1 \), where \( 0 \leq t \leq 1 \). As we revolve this line segment around the \(x\)-axis, it forms a frustum. The radii of the frustum at the ends are the values of \(y\) at \(t = 0\) and \(t = 1\).
02

Determine Radii

Calculate the radii of the frustum using \( y = t + 1 \). For \(t = 0\), \(y = 1\), so \(r_1 = 1\). For \(t = 1\), \(y = 2\), so \(r_2 = 2\). Hence, the radii are \(r_1 = 1\) and \(r_2 = 2\).
03

Derive Slant Height

Determine the slant height of the frustum. The distance between the points \((0,1)\) and \((2,2)\) is the slant height. Use the distance formula: \(slant\ height = \sqrt{(2-0)^2 + (2-1)^2} = \sqrt{5}\).
04

Apply Geometry Formula for Surface Area

According to the geometry formula for the surface area of a frustum, \(\text{Area} = \pi(r_1 + r_2)(\text{slant height})\). Substitute the values: \(\text{Area} = \pi(1 + 2)(\sqrt{5}) = 3\pi\sqrt{5}\).
05

Verification with Calculus

Use the parametric surface area formula for revolving a curve about x-axis: \( A = 2\pi \int_{0}^{1} y \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt \). Substitute \( y = t + 1 \), \( \frac{dx}{dt} = 2 \), and \( \frac{dy}{dt} = 1 \) into the formula: \( A = 2\pi \int_{0}^{1} (t+1) \sqrt{2^2 + 1^2} dt = 2\pi \int_{0}^{1} (t+1)\sqrt{5} dt \).
06

Evaluate the Integral

Calculate the integral: \( A = 2\pi \sqrt{5} \left[ \frac{t^2}{2} + t \right]_{0}^{1} = 2\pi \sqrt{5} \left[ \frac{1}{2} + 1 - 0 \right] = 3\pi \sqrt{5} \).
07

Confirm the Result

Both approaches (geometric formula and calculus) yield the same surface area, \(3\pi\sqrt{5}\). This validates our calculations.

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Key Concepts

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

Surface Area Calculation
Calculating the surface area of a cone frustum involves understanding the structure we are dealing with. A frustum is essentially a cone with its top cut off parallel to the base. Its surface consists of the lateral surface area - without the top and bottom circles. The geometric way to compute the lateral surface area of a frustum is via its radii and slant height. We use the formula:
  • Lateral Surface Area = \( \pi(r_1 + r_2) \times \text{slant height} \)
Here, \( r_1 \) and \( r_2 \) stand for the radii at the ends. The slant height is the hypotenuse of the right triangle formed by the height and difference of the radii. This understanding will help us visualize the frustum in the cone's three-dimensional and geometrical perspective.
Parametrization
Parametrization is a technique in mathematics used to represent a curve. Here, parameters \( (x, y) \) are functions of another variable, denoted as \( t \). The given parametric equations, \( x=2t \) and \( y=t+1 \) where \( 0 \leq t \leq 1 \), describe a straight line segment in the coordinate plane.
  • By revolving this line around the x-axis, we get a 3D shape known as the frustum of a cone.
  • This frustum is bounded by circular ends at the values \( t=0 \) and \( t=1 \), giving the radii of the frustum bases.
Understanding this parametrization allows us to compute dimensions crucial for surface area calculation, such as the radii and the slant height.
Geometry Formula
Utilizing a formula from geometry simplifies finding the surface area of a frustum. This formula, \( \pi(r_1 + r_2)(\text{slant height}) \), leverages the basic principles of geometry. The fundament of this formula is based on understanding the constructed shape:
  • The frustum's shape is akin to a hollow cone with removed top.
  • The total surface area includes only the lateral or side surface area.
The simplicity of this formula lies in using the radii \( r_1 \) and \( r_2 \) and the slant height, which forms the lateral surface as a trapezoid when unfolded. It's effective for quick calculations without needing integral calculus.
Integral Evaluation
Integral evaluation in surface area calculations removes the constraints of simple geometric shapes. Using calculus enables exact calculations for more complex or parametric surfaces. The integral formula for the surface area of a surface generated by revolving a parametric curve around the x-axis is:
  • \( A = 2\pi \int_{0}^{1} y \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt \)
Here, we substitute the parametric functions, yielding the integrand that, upon evaluation, matches the geometry formula. This consistent result from different approaches assures its correctness, highlighting the power of integral calculus to confirm geometric findings.

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Most popular questions from this chapter

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 \(=\pi r(\text { slant height). }\)

Write an integral for the area of the surface generated by revolving the curve \(y=\cos x,-\pi / 2 \leq x \leq \pi / 2,\) about the \(x\) -axis. In Section 8.5 we will see how to evaluate such integrals.

Use the shell method to find the volumes of the solids generated by re- volving the regions bounded by the curves and lines about the \(y\)-axis. \(y=2 x-1, \quad y=\sqrt{x}, \quad x=0\)

Modeling surface area The lateral surface area of the cone swept out by revolving the line segment \(y=x / \sqrt{3}, 0 \leq x \leq \sqrt{3},\) about the \(x\) -axis should be \((1 / 2)(\text { base circumference) (slant height) }=\) \((1 / 2)(2 \pi)(2)=2 \pi .\) What do you get if you use Equation \((8)\) with \(f(x)=x / \sqrt{3} ?\) Graph cannot copy

Constant density Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=x^{3}\) from \(x=0\) to \(x=1\) .
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