/*! 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 42 Suppose \(f\) is a nonnegative f... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose \(f\) is a nonnegative function with a continuous first derivative on \([a, b] .\) Let \(L\) equal the length of the graph of \(f\) on \([a, b]\) and let \(S\) be the area of the surface generated by revolving the graph of \(f\) on \([a, b]\) about the \(x\) -axis. For a positive constant \(C\), assume the curve \(y=f(x)+C\) is revolved about the \(x\) -axis. Show that the area of the resulting surface equals the sum of \(S\) and the surface area of a right circular cylinder of radius \(C\) and height \(L\) 1\. The surface area of the first cone \((200 \sqrt{5} \pi)\) is twice as great as the surface area of the second cone \((100 \sqrt{5} \pi)\) 2\. The surface area is \(63 \sqrt{10} \pi .\) 3. The surface is a cylinder of radius \(c\) and height \(b-a .\) The area of the curved surface is \(2 \pi c(b-a)\)

Short Answer

Expert verified
Answer: The area of surface \(S'\) is equal to the sum of the area of surface \(S\) and the surface area of a right circular cylinder with radius \(C\) and height \(L\).

Step by step solution

01

Find the length of the graph of f

To find the length of the graph of f, we use the arclength formula, which is given by \(L = \int_a^b \sqrt{1 + \left(f'(x)\right)^2} \, dx\).
02

Find the surface area of the surface generated by revolving f(x) about the x-axis

To find the surface area of the original surface, we use the surface area formula for a surface of revolution: \(S = 2\pi \int_a^b f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx\).
03

Find the surface area of the surface generated by revolving f(x) + C about the x-axis

Similarly, we find the surface area of the surface generated by revolving \(f(x)+C\) about the x-axis using the same surface area formula: \(S' = 2\pi \int_a^b [f(x)+C] \sqrt{1 + \left(f'(x)\right)^2} \, dx\).
04

Calculate the difference between S' and S

To prove the statement, we need to find the difference between \(S'\) and \(S\), which can be done by subtracting one integral from the other: \(\Delta S = S' - S\) \(\Delta S = 2\pi \int_a^b [f(x)+C] \sqrt{1 + \left(f'(x)\right)^2} \, dx - 2\pi \int_a^b f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx\).
05

Simplify the difference and find the surface area of a right circular cylinder

We can simplify the difference of the integrals: \(\Delta S = 2\pi \int_a^b C \sqrt{1 + \left(f'(x)\right)^2} \, dx\). Recall that we also want to show that \(\Delta S\) equals the surface area of a right circular cylinder of radius \(C\) and height \(L\). The surface area of a cylinder is \(A_\text{cylinder} = 2 \pi c(b-a) = 2\pi C L.\) To prove the equivalence, it now suffices to show that \(\int_a^b C \sqrt{1 + \left(f'(x)\right)^2} \, dx = CL\).
06

Prove the equivalence

Using the arclength formula from Step 1, we can substitute \(L\) into the integral on the right-hand side: \(\int_a^b C \sqrt{1 + \left(f'(x)\right)^2} \, dx = C \int_a^b \sqrt{1 + \left(f'(x)\right)^2} \, dx = CL\). Therefore, it has been shown that the area of the resulting surface equals the sum of \(S\) and the surface area of a right circular cylinder of radius \(C\) and height \(L\).

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Key Concepts

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

Arclength of a Curve
To find the length of a curve on a graph, we use the concept of arclength. This involves adding up the lengths of local linear segments approximating the curve over a specific interval. The arclength formula is integral-based and is derived from the Pythagorean theorem. If you have a function \( f(x) \) that is continuous and differentiable on an interval \([a, b]\), you can calculate its arclength using the formula:
  • \( L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \)
Here, \(f'(x)\) represents the derivative of \(f\) with respect to \(x\), and it gives us the slope of the curve at any point. The square root part assures that we're considering the actual curved distance rather than just horizontal or vertical changes. Breaking it down this way allows us to convert the curve into tiny straight segments and sum up their lengths to get \(L\), the total arclength.
Integral Calculus
Integral calculus is a branch of mathematics that deals with the concept of integration. Integration is essentially the reverse process of differentiation. It's used to find things like areas under curves, volumes, and many other quantities. With respect to the surface area of revolution and arclength, integral calculus allows us to calculate the total sum of infinitely small quantities that make up the whole.
When trying to find areas with a given function \(f(x)\), you're often looking to solve an integral over a specific interval. The basic structure of an integral, like
  • \( \int_a^b f(x) \,dx \)
effectively sums an infinite number of infinitesimally small strips, from \(x = a\) to \(x = b\). The impact of integral calculus extends beyond just simple areas and lengths, stretching into dimensions like calculating the surface area of shapes created by revolving functions about an axis.
Right Circular Cylinder
A right circular cylinder is a simple geometric figure commonly encountered in geometry. Imagine a tin can with a circular base and a fixed height. In mathematical terms, a right circular cylinder has:
  • Two parallel circular bases, having the same radius \(r\).
  • A height \(h\), which is the distance between the two bases.
  • The axis of the cylinder, which connects the centers of these two circular bases.
If you need to find the lateral or 'side' surface area of this cylinder, you multiply the circumference of the circular base (\(2\pi r\)) by its height \(h\). Therefore, the formula for the lateral surface area of a right circular cylinder is
  • \( A = 2\pi r h \).
Understanding this shape and its properties is particularly important when dealing with surface areas and revolutions in calculus.
Surface Area Formula
The surface area formula is crucial in the context of revolution and involves integrating along a path to compute the area "swept out" when a curve revolves around an axis. When dealing with functions like \( f(x) \), you find the surface area of the shape formed by rotating this curve around the x-axis using:
  • \( S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \, dx \).
This formula factors in both the length of the segments that make up the curve and their radius from the axis of rotation.
When additional components are introduced, such as an additional constant \( C \) that transforms \( f(x) \) into \( f(x) + C \), these elements impact the surface by adding the area contributed by a cylinder. Thus, the full understanding of this formula allows us to comprehend how simply adding a constant can mean including more geometrical area, specifically making the surface area equivalent to adding a cylindrical surface of specified dimensions.

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