Chapter 6: Q. 71 (page 540)

$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \sin x, [a, b] = [0, π]$

Short Answer

Expert verified

$The surface area of the solid of revolution obtained by revolving f (x) = sinx around the x - axis from x = 0 to x = π is : S = 2 π \sqrt{2} + π \ln (\sqrt{2} + 1) - π \ln (\sqrt{2} - 1)$

Step by step solution

Step 1. Given Information

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

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

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

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$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \sin x, [a, b] = [0, π]$

Short Answer

Step by step solution

Step 1. Given Information

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

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InExercises63–72,setupandsolveadefiniteintegraltofindtheexactareaofeachsurfaceofrevolutionobtainedbyrevolvingthecurvey=f(x)aroundthex-axisontheinterval[a,b].f(x)=sinx,[a,b]=[0,π]

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

Applied Mathematics

Logic and Functions

Theoretical and Mathematical Physics

Statistics

Calculus

Probability and Statistics

Study anywhere. Anytime. Across all devices.

$In Exercises 63 - 72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f (x) around the x - axis on the interval [a, b] . f (x) = \sin x, [a, b] = [0, π]$