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Chapter 6: Problem 4

Assume \(g\) is a nonnegative function with a continuous first derivative on \([c, d] .\) The curve \(x=g(y)\) on \([c, d]\) is revolved about the y-axis. Explain how to find the area of the surface that is generated.

Short Answer

Expert verified
Question: Find the area of the surface generated by revolving the curve x = g(y) in the interval [c,d] around the y-axis. Answer: To find the surface area, integrate the following expression over the interval [c,d]: \(A = \int_{c}^{d} 2\pi g(y) \sqrt{g'(y)^2 + 1} dy\)

Step by step solution

01

1. Determine the differential element of surface area

Consider a small segment of the curve as a right triangle with base length equal to the infinitesimal change in x, \(dx\), and altitude length equal to the infinitesimal change in y, \(dy\). By the Pythagorean theorem, the hypotenuse length, \(ds\), is given by: \(ds = \sqrt{dx^2 + dy^2}\) We know that \(x = g(y)\), thus, \(dx = g'(y) dy\), where \(g'(y)\) is the derivative of g(y) with respect to y. Now, substitute this into the expression for ds: \(ds = \sqrt{(g'(y) dy)^2 + dy^2}\)
02

2. Set up the integral for the surface area

To find the surface area, we will multiply the differential element, \(ds\), by the circumference of the circle formed due to the rotation, \(2\pi x\). Then we will integrate over the interval [c,d]: \(A = \int_{c}^{d} 2\pi x ds\) Now, substitute the expressions for \(x\) and \(ds\) in terms of y: \(A = \int_{c}^{d} 2\pi g(y) \sqrt{(g'(y) dy)^2 + dy^2}\)
03

3. Simplify and solve the integral

To simplify this expression, we can first factor out a \(dy^2\) term from the square root: \(A = \int_{c}^{d} 2\pi g(y) \sqrt{dy^2 (g'(y)^2 + 1)}\) Now, cancel the \(dy^2\) term inside the square root with the one outside: \(A = \int_{c}^{d} 2\pi g(y) \sqrt{g'(y)^2 + 1} dy\) Finally, solve the integral to obtain the surface area generated by revolving the curve x=g(y) on the interval [c,d] around the y-axis.

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Key Concepts

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

Integral Calculus
Integral calculus is an essential tool in mathematics used to find things like areas under curves, rates of accumulation, and, as in our exercise, surface areas of revolved curves. The integral is a mathematical operation used to combine tiny pieces of data鈥攕uch as the segments of a curve鈥攊nto a whole picture.
When analyzing the surface of revolution problem, we use definite integrals to sum up infinitesimal calculations over an interval
  • The integral \((A = \int_{c}^{d} 2\pi x ds\)) represents adding up small elements to form a complete surface area.
  • Each tiny piece belongs to a curve segment that revolves around an axis to form a larger surface.
Using calculus, mathematicians can accurately calculate complex shapes and areas that are difficult with simple geometry.
Pythagorean Theorem
The Pythagorean Theorem is a simple yet powerful concept from geometry. It states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides \( (a^2 + b^2 = c^2) \).
In finding the surface area of revolution, this theorem helps calculate the arc length (or differential element) of a curve segment.
The arc length is determined using:
  • The infinitesimal change in x \( (dx) \) and y \( (dy) \), which form the base and height of a right triangle.
  • The hypotenuse \( (ds) \) becomes the actual curve length between two infinitesimally close points.
  • By equation, \(ds = \sqrt{dx^2 + dy^2}\), which is grounded in the Pythagorean Theorem.
This calculation helps in breaking down curve segments accurately before integrating them.
Differential Element
The differential element is a small part of a larger whole, a concept crucial in calculus for approximations and integration. It refers to an infinitesimally small measurement of a quantity鈥攕uch as length, area, or volume.
In the context of this problem, the differential element of surface area comes from the arc length \( ds \) calculated using the Pythagorean theorem.
This small 'slice' of the curve contributes to calculating total surface area by:
  • Measuring the length of a tiny segment of the curve.
  • Multiplying it by the circumference due to rotation \( (2\pi x) \).
  • Adding all these elements using integration over a given interval.
Understanding differential elements allows for precise calculation of surfaces that arise from rotating curves.
Curve Rotation
Curve rotation involves creating a three-dimensional shape by revolving a two-dimensional curve around an axis. This action simplifies to a known form鈥攍ike a cylinder or cone鈥攂ased on the curve's properties and the direction of rotation.
For the problem mentioned, the revolution happens about the y-axis, generating a complex surface known as a surface of revolution. This surface is:
  • Created by rotating the curve given by \(x = g(y)\).
  • Analyzed by considering the infinitesimally small surface area contributions along the curve.
  • Evaluated by integrating these contributions over the specified interval \([c,d]\).
Rotation leads to applications in engineering and design where complex surfaces are typical, making understanding curve rotation and integration vital.

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Most popular questions from this chapter

Surface area using technology Consider the following curves on the given intervals. a. Write the integral that gives the area of the surface generated when the curve is revolved about the given axis. b. Use a calculator or software to approximate the surface area. \(y=e^{x},\) for \(0 \leq x \leq 1 ;\) about the \(y\) -axis

Work by two different integrals A rigid body with a mass of 2 kg moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{m}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{1}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).

Determine whether the following statements are true and give an explanation or counterexample. a. When using the shell method, the axis of the cylindrical shells is parallel to the axis of revolution. b. If a region is revolved about the \(y\) -axis, then the shell method must be used. c. If a region is revolved about the \(x\) -axis, then in principle, it is possible to use the disk/washer method and integrate with respect to \(x\) or to use the shell method and integrate with respect to \(y\)

A torus (doughnut) Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral. A torus (doughnut) Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

Emptying a conical tank An inverted cone (base above the vertex) is \(2 \mathrm{m}\) high and has a base radius of \(\frac{1}{2} \mathrm{m} .\) If the tank is full, how much work is required to pump the water to a level \(1 \mathrm{m}\) above the top of the tank?
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