Answer to Question #198254 in Calculus for Ben

Question

Area of Surface of Revolution

1. 9𝑦 = 𝑥^2 , 0 ≤ 𝑥 ≤ 2 ; x-axis

2. 𝑦 = 𝑙𝑛(𝑥^2 − 1) , 2 ≤ 𝑥 ≤ 3 ; y-axis

3. 𝑦 = 3^√𝑥 , 0 ≤ 𝑥 ≤ 1 ; y-axis

4. 𝑦 = √9 − 𝑥^2 , −2 ≤ 𝑥 ≤ 2 ; x-axis

Answer 1

1. 9𝑦 = 𝑥² , 0 ≤ 𝑥 ≤ 2 ; x-axis ..............Equation(1)

Area of the surface is given by

A = $\int$ ₀² 2 $*\pi$ $*$ y $\sqrt{1 + (\dfrac{dy}{dx})^2}$ dx

Differentiating Equation(1) with respect to x, we have

$\dfrac{dy}{dx}$ = 2x / 9

So, area is

A = $\int$ ₀² 2$*\pi$ $*$ $\dfrac{x^2}{9}$ $\sqrt{1 + \dfrac{4x^2}{81}}$ dx ........................Equation(2)

A = $\dfrac{4}{81}$ $\pi$ $\int$ ₀² x²$\sqrt{x^2 + \dfrac{81}{4}}$ dx

Let x = $\dfrac{9}{2}\tan\theta$

dx = (9/2) sec² $\theta$ d$\theta$

A = $\dfrac{4}{81}$ $\pi$ $\int$ ₀ $\phi$ $\dfrac{81}{4} tan^2\theta$ $\dfrac{9}{2}\sec\theta$ (9/2) sec² $\theta$ d$\theta$ ..............[ $\phi$ = arctan(4/9)]

A = $\pi$ $\int$ ₀ $\phi$ $\ tan^2\theta$ $\dfrac{9}{2}\sec\theta$ $\dfrac{9}{2}$ sec² $\theta$ d$\theta$

A = $\dfrac{81\pi}{4}$ $\int$₀ $\phi$ $tan^2\theta sec^3\theta$ d$\theta$

On solving the above integral we get

A = 2.43 square units.

2. 𝑦 = 𝑙𝑛(𝑥^2 − 1) , 2 ≤ 𝑥 ≤ 3 ; y-axis ...............Equation(2)

x varies from 2 to 3

So, y varies from ln 3 to ln 8

Area of the surface is given by,

A = $\int$ _{ln 3} ^{ln 8} 2 $*\pi$ $*$ x $\sqrt{1 + (\dfrac{dx}{dy})^2}$ dy

Differentiating Equation (2) with respect to y we have

$\dfrac{dx}{dy} = \dfrac{e^y}{2\sqrt{e^y + 1}}$

A = $\int$ _{ln 3} ^{ln 8} 2 $*\pi$ $*$ $\sqrt{e^y + 1}$ $*$ $\dfrac{e^y + 1}{2\sqrt{e^y + 1}}$ dy

A = $\int$ _{ln 3} ^{ln 8} $*\pi$ (e^y + 1 )dy

A = $\pi$ [(e^y + y ] _{ln 3} ^ln8

A = 18.78 square units.

3. 𝑦 = 3^√𝑥 , 0 ≤ 𝑥 ≤ 1 ; y-axis ................Equation(3)

x = y³

dx/dy = 3y²

The area is given as

A = $\int$ ₀ ¹ 2 $*\pi$ $*$ x $\sqrt{1 + (\dfrac{dx}{dy})^2}$ dy

A = $\int$ ₀ ¹ 2 $\pi$ y³ $\sqrt{1 + 9y^4}$ dy

Let 1 + 9y⁴ = t

36 y³ dy = dt

A = $\int$ ₁ ¹⁰ 2 $\pi$ $\dfrac{\sqrt{t}}{36}$ dt

A = $\dfrac{2\pi}{36} * \dfrac{2}{3} \int$ ₁ ¹⁰ [ t^1.5 ]₁ ¹⁰

A = 3.561 square units.

4. 𝑦 = √9 − 𝑥^2 , −2 ≤ 𝑥 ≤ 2 ; x-axis ...........................Equation(4)

Area of the surface is given by

A = $\int$ _-2 ² 2 $*\pi$ $*$ y $\sqrt{1 + (\dfrac{dy}{dx})^2}$ dx

Differentiating Equation(1) with respect to x, we have

$\dfrac{dy}{dx}$ ​= $\dfrac{- x}{\sqrt{9 -x^2}}$

So, area is

A = $\int$ _-2 ² 2 $*\pi$ $*$ $\sqrt{9-x^2} * \dfrac{3}{\sqrt{9 - x^2}}$ dx

A = $\int$ _-2 ² 6$\pi$ dx

A = 24 $\pi$ square units

A = 75.36 square units.