   Chapter 10.5, Problem 65E

Chapter
Section
Textbook Problem

# Finding the Area of a Surface of Revolution In Exercises 65-68, find the area of the surface formed by revolving the polar equation over the given interval about the given line.Polar Equation Interval Axis of Revolution r = e a θ 0 ≤ θ ≤ π 2 θ = π 2

To determine

To calculate: The value of the surface area formed by revolving the polar equation r=eaθ over the given interval 0θπ2 about the line.

Explanation

Given:

The polar equation is given r=eaθ and the interval 0θπ2 axis of revolution is about the line θ=π2.

Formula Used:

s=2παβf(θ)cosθ(f(θ))2+(f(θ))2dθ

Calculation:

Given polar equation is r=eaθ where interval is 0θπ2 and axis of revolution is polar axis.

Now, area of surface will be given by below formula;

s=2παβf(θ)cosθ(f(θ))2+(f(θ))2dθ

Here, r=eaθ, α=0, β=π2

And

drdθ=f(θ)=aeaθ

Substitute these values in the above formula and get;

s=2παβf(θ)cosθ(f(θ))2+(f(θ))2dθ=2π0π2eaθcosθ(eaθ)2+(aeaθ)2dθ=2π0π2eaθcosθe2aθ(1+a2)dθ=2π1+a20π2e2aθcosθdθ

Now further solve and get,;

Let I=e2aθcosθdθ

Apply integration by part,

I=e2aθcosθdθ(ddx(e2aθ)cosθdθ)dθ=e2aθsinθ

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