Chapter 6: Q. 72 (page 540)

$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \cos x, [a, b] = [- \frac{π}{2}, \frac{π}{2}]$

Short Answer

Expert verified

$Therefore, the surface area of the solid of revolution obtained by revolving f (x) = \cos x around the x - axis from x = - \frac{π}{2} to x = \frac{π}{2} is : 2 π \sqrt{2} + 2 π \ln (\sqrt{2} + 1)$

Step by step solution

Step 1. Given Information

The curve, $f (x) = \cos x$

Step 2. Finding the surface area of the solid of revolution

$The surface area of the solid of revolution obtained by revolving f (x) around the x - axis from x = a to x = b is : 2 π \int_{a}^{b} f (x) \sqrt{1 + (f' {(x))}^{2}} d x The derivative of f (x) = \cos x is f' (x) = - \sin x Therefore, the surface area of the solid of revolution obtained by revolving f (x) around the x - axis from x = - \frac{π}{2} to x = \frac{π}{2} is : 2 π \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos x \sqrt{1 + {(- \sin x)}^{2}} d x Put, \sin x = a, \cos x d x = d a Therefore, surface area, S = 2 π \times 2 \int_{0}^{1} \sqrt{1 + a^{2}} d a S = 4 π \int_{0}^{1} \sqrt{1 + a^{2}} d a S = 4 π [{\frac{a}{2} \sqrt{1 + a^{2}} + \frac{1}{2} \ln |a + \sqrt{1 + a^{2}}|}_{0}^{1}] S = 4 π [(\frac{1}{2} \sqrt{1 + 1^{2}} - \frac{0}{2} \sqrt{1 + 0^{2}}) + \frac{1}{2} \ln |1 + \sqrt{1 + 1^{2}}| - \frac{1}{2} \ln |0 + \sqrt{1 + 0^{2}}|] S = 4 π [\frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\sqrt{2} + 1)] S = 2 π \sqrt{2} + 2 π \ln (\sqrt{2} + 1)$

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \cos x, [a, b] = [- \frac{π}{2}, \frac{π}{2}]$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the surface area of the solid of revolution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Applied Mathematics

Statistics

Probability and Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

InExercises63–72,setupandsolveadefiniteintegraltofindtheexactareaofeachsurfaceofrevolutionobtainedbyrevolvingthecurvey=f(x)aroundthex-axisontheinterval[a,b].f(x)=cosx,[a,b]=−π2,π2

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the surface area of the solid of revolution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Applied Mathematics

Statistics

Probability and Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \cos x, [a, b] = [- \frac{π}{2}, \frac{π}{2}]$