/*! 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 8 Computing surface areas Find the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Computing surface areas Find the area of the surface generated when the given curve is revolved about the given axis. \(y=12-3 x,\) for \(1 \leq x \leq 3 ;\) about the \(x\) -axis

Short Answer

Expert verified
Answer: The surface area generated is \(72\pi\) square units.

Step by step solution

01

Find the derivative of y

(In this case \(y' = -3\)) \(y(x) = 12 - 3x\) To find the derivative, we apply the power rule to each term: \(y'(x) = \frac{d}{dx} (12 - 3x)= -3\)
02

Set up the integral for surface area

Now we're ready to plug into the surface area formula. Let's use our information into the formula and set up the integral for the surface area. Here, we just substitute the expression for the function \(y\), the interval and the derivative \(y'\): \(A = 2\pi \int_1^3 (12-3x) \sqrt{1 + (-3)^2} dx\) Now, simplify the integral: \(A = 6\pi \int_1^3 (12-3x) dx\)
03

Calculate the integral to find A

Now we'll evaluate the integral by finding the antiderivative of \((12-3x)\) and then applying the Fundamental Theorem of Calculus: The antiderivative of \((12-3x)\) with respect to x is: \(12x - \frac{3}{2}x^2 + C\) Now we'll use the Fundamental Theorem of Calculus: \(A = 6\pi [(12(3) - \frac{3}{2}(3)^2) - (12(1) - \frac{3}{2}(1)^2)]\) \(A = 6\pi [(36 - 13\frac{1}{2}) - (12 - 1\frac{1}{2})]\) \(A = 6\pi [22\frac{1}{2} - 10\frac{1}{2}]\) \(A = 6\pi [12]\) \(A = 72\pi\) The surface area generated when the curve \(y = 12 - 3x\) is revolved about the \(x\)-axis for the interval \(1 \leq x \leq 3\) is \(72\pi\) square units.

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.

Integral Calculus
Integral calculus, a fundamental part of mathematics, is concerned with the process of finding the area under curves and is the reverse operation of differentiation. In essence, it allows us to compute total size or value, such as areas under a curve, displacement given a velocity function, or the volume of a solid of revolution.

When we talk about solving problems involving curves and areas, we use the term 'integral' to refer to the algebraic sum of the areas of rectangles that approximate the region under a curve. As the width of those rectangles approaches zero, this approximation gets infinitely close to the true area, resulting in a definite integral for a specific interval.

Integral calculus also helps us solve more complex real-world applications, such as finding the growth of populations, determining the amount of material needed for a construction project, and in our case, computing the surface area of a 3D object formed by revolving a 2D curve around an axis.
Surface Area Integral
When a curve is rotated about an axis, it sweeps out a three-dimensional shape known as a solid of revolution. The surface area integral is a specialized tool in integral calculus used to calculate the curved surface area of such objects.

The general formula for finding the surface area \(A\) of a solid revolved around the x-axis is given by: \[ A = 2\pi \int_{a}^{b} y(x) \sqrt{1 + [y'(x)]^2} \, dx \] where \(y(x)\) is the function defining the curve, \(a\) and \(b\) are the limits of integration representing the interval on the x-axis, and \(y'(x)\) is the derivative of \(y\) with respect to \(x\). This integral includes both the curve's radius from the axis of rotation and the length of an infinitesimally small arc of the curve—key components for calculating surface area.

To find the solution, you would typically set up the integral as the formula indicates and carry out the integration process to reach an expression for the total surface area.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the process of differentiation and the process of integration in a beautiful, practical symmetry. It has two main parts, and both are essential for solving calculus problems involving definite integrals.

The first part states that if a function is continuous over an interval and is the integral of its derivative on that interval, then the function's integral over that interval is equal to the difference in its end values. This is formally expressed as: \[ F(b) - F(a) = \int_{a}^{b} f'(x) \, dx \] where \(F\) is the antiderivative of \(f'\).

The second part helps us compute definite integrals without summing infinite approximations. It suggests that if \(F\) is an antiderivative of \(f\) over an interval \[a, b\], then the definite integral of \(f\) is \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]. This theorem is particularly crucial when calculating areas and volumes, as it offers a method to evaluate definite integrals easily, which represent these quantities.

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

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _______. $$v(t)=2 t+6,\( for \)0 \leq t \leq 8$$

A glass has circular cross sections that taper (linearly) from a radius of 5 cm at the top of the glass to a radius of 4 cm at the bottom. The glass is 15 cm high and full of orange juice. How much work is required to drink all the juice through a straw if your mouth is 5 cm above the top of the glass? Assume the density of orange juice equals the density of water.

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. $$y=\cos x, \text { for } 0 \leq x \leq \frac{\pi}{2} ; \text { about the } x \text { -axis }$$

At noon \((t=0)\). Alicia starts running along a long straight road at \(4 \mathrm{mi} / \mathrm{hr}\). Her velocity decreases according to the function \(v(t)=\frac{4}{t+1},\) for \(t \geq 0 .\) At noon, Boris also starts running along the same road with a 2 -mi head start on Alicia: his velocity is given by \(u(t)=\frac{2}{t+1},\) for \(t \geq 0 .\) Assume \(t\) is measured in hours. a. Find the position functions for Alicia and Boris, where \(s=0\) corresponds to Alicia's starting point. b. When, if ever, does Alicia overtake Boris?

Challenging surface area calculations Find the area of the surface generated when the given curve is revolved about the given axis. \(y=x^{3 / 2}-\frac{x^{1 / 2}}{3},\) for \(1 \leq x \leq 2 ;\) about the \(x\) -axis

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.