/*! 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 7 Find the area of the surface gen... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=8 \sqrt{x} \text { on }[9,20]$$

Short Answer

Expert verified
The formula for the surface area of revolution when revolving a curve about the x-axis is: $$S = 2\pi \int_a^b y \sqrt{1 + (\frac{dy}{dx})^2} dx$$

Step by step solution

01

Find the derivative of the curve with respect to x

Differentiate \(y=8\sqrt{x}\) with respect to \(x\): $$\frac{dy}{dx} = \frac{d}{dx}(8\sqrt{x}) = 8(\frac{d}{dx}\sqrt{x}) = 8(\frac{1}{2}x^{-1/2}) = 4x^{-1/2}$$
02

Substitute the function and its derivative into the formula

Now substitute \(y=8\sqrt{x}\) and \(\frac{dy}{dx}=4x^{-1/2}\) into the surface area formula: $$S = 2\pi \int_9^{20} 8\sqrt{x} \sqrt{1 + (4x^{-1/2})^2} dx$$
03

Simplify the expression under the square root

Simplify the expression inside the square root: $$1 + (4x^{-1/2})^2 = 1 + 16x^{-1}$$ Now the integral becomes: $$S = 2\pi \int_9^{20} 8\sqrt{x} \sqrt{1 + 16x^{-1}} dx$$
04

Make a substitution

Let's make a substitution to simplify the integral. Let \(u=x^{\frac{1}{2}}\), so \(x=u^2\) and \(dx=2u du\). The new integration limits are \(u=\sqrt{9}=3\) and \(u=\sqrt{20}\). Now the integral becomes: $$S = 2\pi \int_3^{\sqrt{20}} 8u^2 \sqrt{1 + 16u^{-2}} (2u) du$$
05

Simplify and solve the integral

Simplify the integrand: $$S = 32\pi \int_3^{\sqrt{20}} u^3 \sqrt{1 + 16u^{-2}} du$$ Now, solve the integral: $$S = 32\pi \left[ \frac{1}{5}u^5\sqrt{1+16u^{-2}} - 4\ln(u+\sqrt{1+16u^{-2}}) \right]_3^{\sqrt{20}}$$
06

Evaluate the definite integral

Evaluate the definite integral using the antiderivative and the limits of integration: $$S = 32\pi \left[\left(\frac{1}{5}(\sqrt{20})^5\sqrt{1+16(\sqrt{20})^{-2}} - 4\ln(\sqrt{20}+\sqrt{1+16(\sqrt{20})^{-2}})\right) - \left(\frac{1}{5}(3)^5\sqrt{1+16(3)^{-2}} - 4\ln(3+\sqrt{1+16(3)^{-2}})\right)\right]$$ After evaluating the expression above, we will find the area of the surface generated when the curve is revolved about the \(x\)-axis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Surface Integral
In the context of finding the area of a surface generated by revolving a curve around an axis, the concept of a surface integral is employed. This isn't like other integrals that simply sum the heights or areas; rather, a surface integral sums over a surface area in space. This is used to calculate complex surfaces that might arise from rotating a curve, allowing us to account for variations along the surface.

To compute this, you need both the original function and its derivative. Why? Because the derivative helps account for the slope and orientation of the curve as it's revolved. This leads to our integral needing the expression \( \sqrt{1+(\frac{dy}{dx})^2} \), which forms part of the surface area formula. This integral is set up to accommodate the change in area as the curve wraps around the axis of revolution. This method not only demonstrates the power of calculus but also equips you with a tool for handling complex surface calculations in both mathematics and applied science.
Definite Integral
A definite integral represents the total accumulation of quantities, like areas or volumes, over a given interval. In our example, it specifically concerns calculating the area from one point on the curve to another. This method ensures that we account for each infinitesimal piece over the interval \([9, 20]\).
  • Start with finding the derivative needed in the integral.
  • Use it with the original function within the integral limits.
  • Finalize by evaluating to receive a number that represents the area.
These integrals are not just hopeful guesses—they give precise values when properly evaluated, making them reliable tools in both pure and applied mathematics. Integrals take into account each segment of the curve to provide total measurements over specific intervals, bridging the abstract family of calculus with real-world applications.
Curve Revolution Formula
The curve revolution formula is central to problems involving curves spun around an axis. This isn't just any formula—it's specifically designed to compute the surface area of these revolving shapes.

For a curve defined by \(y=f(x)\) being revolved around the \(x\)-axis, the formula is:
  • \(S = 2\pi \int_a^b y \sqrt{1 + (\frac{dy}{dx})^2} \, dx\)

This formula integrates the distance a point on the curve travels to cover the whole surface, thus forming a comprehensive method to derive the surface area from the curve's revolution.

By substituting in our specific function and limits, we apply the formula to achieve a solution tailored to our problem’s needs. This demonstrates how powerful calculus is—not only does it abstractly define shapes, but it also gives us the ability to measure something as seemingly complex as a spun or 'revolved' curve with pinpoint precision. This critical formula is thus both a theoretical and practical marvel, cementing its place as an essential tool in mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

There are several ways to express the indefinite integral of \(\operatorname{sech} x\). a. Show that \(\left.\int \operatorname{sech} x d x=\tan ^{-1}(\sinh x)+C \text { (Theorem } 6.9\right)\) (Hint: Write sech \(x=\frac{1}{\cosh x}=\frac{\cosh x}{\cosh ^{2} x}=\frac{\cosh x}{1+\sinh ^{2} x}\) and then make a change of variables.) b. Show that \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C .\) (Hint: Show that \(\operatorname{sech} x=\frac{\operatorname{sech}^{2} x}{\sqrt{1-\tanh ^{2} x}}\) and then make a change of variables.) c. Verify that \(\int \operatorname{sech} x d x=2 \tan ^{-1} e^{x}+C\) by proving \(\frac{d}{d x}\left(2 \tan ^{-1} e^{x}\right)=\operatorname{sech} x.\)

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x.\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \frac{2 \pi d}{\lambda}} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}.\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

A right circular cylinder with height \(R\) and radius \(R\) has a volume of \(V_{C}=\pi R^{3}\) (height \(=\) radius). a. Find the volume of the cone that is inscribed in the cylinder with the same base as the cylinder and height \(R\). Express the volume in terms of \(V_{C}\). b. Find the volume of the hemisphere that is inscribed in the cylinder with the same base as the cylinder. Express the volume in terms of \(V_{C}\).

Prove that the doubling time for an exponentially increasing quantity is constant for all time.

Sketch a solid of revolution whose volume by the disk method is given by the following integrals. Indicate the function that generates the solid. Solutions are not unique. a. \(\int_{0}^{\pi} \pi \sin ^{2} x d x\) b. \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.