   Chapter 10.2, Problem 64E

Chapter
Section
Textbook Problem

# Graph the curve x = 2 cos θ − cos 2 θ y = 2 sin θ − sin 2 θ If this curve is rotated about the x-axis, find the exact area of the resulting surface. (Use your graph to help find the correct parameter interval.)

To determine

To find: the exact area of the resulting surface if the curve is rotate about x axis for the parametric equation  x=2cosθcos2θ and y=2sinθsin2θ .

Explanation

Given:

The parametric equation for the variable x is as below.

x=2cosθcos2θ (1)

The parametric equation for the variable y is as below.

y=2sinθsin2θ (2)

Calculation:

Substitute (0) for θ in equation (1),

x=2cosθcos2θ=2cos(0)cos2(0)=21=1

Substitute (0) for θ in equation (2),

y=2sinθsin2θ=2sin(0)sin2(0)=0

The values of x and y for each step value of θ t is tabulated in the below table.

 Degrees Radians (θπ180) x y 0 0 1 0 10 0.174611111 1.0299492 0.00528327 20 0.349222222 1.1134349 0.041307007 30 0.523833333 1.2322225 0.134146372 40 0.698444444 1.3586546 0.301138132 50 0.873055556 1.4593943 0.547919655

Graph:

Graph is plotted with the parameter ranging from 0 to 2π .

The surface of revolution is generated twice.

Integration is done with the parameter ranging from 0 to π .

Differentiate the variable x with respect to θ .

x=2acosθ+cos2θdxdθ=2asinθ+sin2θ

Differentiate the variable y with respect to θ :

y=2asinθsin2θdxdθ=2cosθ2cos2θ

The surface area of the surface obtained by rotating curve about the x axis.

S=0π2πy(dxdθ)2+(dydθ)2dθ

The value θ ranges from 0 to π .

Substitute (2asinθ+sin2θ) for dxdt and (2cosθ2cos2θ) for dydt in the above equation.

S=0π2πy(dxdθ)2+(dydθ)2dθ=0π2π(2sinθsin2θ)(2asinθ+sin2θ)2+(2cosθ2cos2θ)2dθ=4π0π(2sinθsin2θ)(2asinθ+sin2θ)2+(2cosθ2cos2θ)2dθ=4π0π(2sinθsin2θ)sin22θ+si

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