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Assume that \(y=f(x)\) is a smooth curve on the interval \([a, b]\) and assume that \(f(x) \geq 0\) for \(a \leq x \leq b\). Derive a formula for the surface area generated when the curve \(y=f(x), a \leq x \leq b\), is revolved about the line \(y=-k(k>0)\)

Short Answer

Expert verified
Use surface area formula: \( S = 2\pi \int_a^b (f(x) + k) \sqrt{1 + (f'(x))^2} \, dx \).

Step by step solution

01

Understand the Problem

We are given a curve described by the function \( y = f(x) \) on the interval \( [a, b] \), and we want to find the formula for the surface area of the solid generated when this curve is revolved around the line \( y = -k \), where \( k > 0 \).
02

Identify Parameters for Revolution

When a curve \( y = f(x) \) is revolved about a horizontal line \( y = -k \), the radius of revolution for a point \( (x, f(x)) \) is the distance from \( f(x) \) to \( -k \), which is \( f(x) + k \).
03

Use the Surface Area Formula for Revolution

The formula for the surface area \( S \) when revolving a curve \( y = f(x) \) about a horizontal axis is given by: \[ S = 2\pi \int_a^b R(x) \sqrt{1 + (f'(x))^2} \, dx \] where \( R(x) = f(x) + k \).
04

Substitute Radius into the Integral

Substitute \( R(x) = f(x) + k \) into the surface area formula: \[ S = 2\pi \int_a^b (f(x) + k) \sqrt{1 + (f'(x))^2} \, dx \]
05

Conclusion of Formula Derivation

We have derived the formula for the surface area of the solid generated by revolving the curve \( y = f(x) \) around the line \( y = -k \): \[ S = 2\pi \int_a^b (f(x) + k) \sqrt{1 + (f'(x))^2} \, dx \]

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Key Concepts

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

Surface Area of Revolution
The surface area of revolution is a fascinating concept in calculus. When you revolve a curve around a line, you create a 3D shape. The surface area of this shape is what we're trying to calculate. To imagine this, picture spinning a curve like y = f(x) around a horizontal line like y = -k. The result is a "ring" or "shell" shape that wraps around that line.

We need an integral to calculate this surface area. Here's the formula for surface area when a curve is revolved:
  • The general formula is:\[ S = 2 \pi \int_a^b R(x) \sqrt{1 + (f'(x))^2} \, dx \]
  • R(x) is the radius of revolution. For our situation, it is given by: \[ R(x) = f(x) + k \]
  • The \( f'(x) \) is the derivative of the function. It represents the slope and helps in calculating the distance along the curve.
Understanding the formula can help you solve similar problems where you revolve curves around horizontal or vertical lines. It's about understanding the shape you're making in 3D and how to measure its skin.
Integration
Integration is a powerful tool in calculus used to calculate areas, volumes, and more. In the context of surface area of revolution, integration helps us sum up small surface bands to find the total surface area.

The integral you see in the surface area formula performs this task. It adds up infinitely small portions along the curve from point 'a' to point 'b'. This gives us the total surface area of the shape we created.

  • The integral \( \int_a^b \) signifies that we're evaluating from x = a to x = b along the x-axis.
  • Inside the integral, \( \sqrt{1 + (f'(x))^2} \) ensures that we consider the curve's slope impact on the total length.
By understanding integration, you're equipped to tackle a wide range of calculus problems involving areas and volumes. It's all about adding up small parts to find a complete solution.
Differentiation
Differentiation is another core principle of calculus. It's the process of finding the derivative, which measures how a function changes at any given point. In the problem of surface area of revolution, \( f'(x) \) is crucial. It tells us the slope of the curve at every point along \([a, b]\).

This slope information is essential because it helps us understand how the curve stretches. When calculating surface areas, knowing how steep a curve is can change how much surface area it has when revolved around a line.

Working with derivatives involves:
  • Applying differentiation rules to find \( f'(x) \) from \( f(x) \).
  • Using the derivative in calculations to ensure accurate area measurements.
Differentiation gives us the tools to measure the change. By understanding these tools, finding slopes and curve behavior becomes intuitive, aiding tremendously in complex calculus problems.

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Most popular questions from this chapter

Suppose that a hollow tube rotates with a constant angular velocity of \(\omega \mathrm{rad} / \mathrm{s}\) about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0\), and assume that the object is at rest and \(r=0\) when \(t=0\). It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$ r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)] $$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) (a) Graph \(r\) versus \(t\) for \(0 \leq t \leq 1\). (b) Assuming that the tube has a length of \(1 \mathrm{~m}\), approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate \(d r / d t\) at the instant that the object reaches the end of the tube.

(a) Find the force (in \(\mathrm{N}\) ) on the deck of a sunken ship if its area is \(160 \mathrm{~m}^{2}\) and the pressure acting on it is \(6.0 \times 10^{5} \mathrm{~Pa}\) (b) Find the force (in \(\mathrm{lb}\) ) on a diver's face mask if its area is \(60 \mathrm{in}^{2}\) and the pressure acting on it is \(100 \mathrm{lb} / \mathrm{in}^{2}\).

Determine whether the statement is true or false. Explain your answer. The equation \(\cosh x=\sinh x\) has no solutions.

Use a CAS to find the exact area of the surface generated by revolving the curve about the stated axis. \(8 x y^{2}=2 y^{6}+1,1 \leq y \leq 2 ; y\) -axis

Evaluate the integrals. $$ \int_{0}^{1 / 2} \frac{d x}{1-x^{2}} $$

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