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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

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Section
BuyFindarrow_forward

Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Finding the Area of a Polar Region In Exercises 19-26, use a graphing utility to graph the polar equation. Find the area of the given region analytically.

Between the loops of   r = 1 + 2 cos θ

To determine

To calculate: The value of the area between the loops of polar equation r=1+2cosθ and draw it with the use of graphing utility.

Explanation

Given:

The polar equation r=1+2cosθ.

Formula used:

The area of the polar equation is given by:

A=12αβ[f(θ)]2dθ

Where, α and β are limits of the integration.

Calculation:

Consider the polar equation r=1+2cosθ.

Now, use the following steps in the TI-83 calculator to obtain the graph:

Step 1: Press ON button to open the calculator.

Step 2: Press MODE button and then scroll down to press pol and press ENTER button.

Step 3: Now, press the button Y= and enter the provided equation.

Step 4: Press WINDOW button and then set the window as follows:

Xmin=1,Xmax=4,Ymin=2 and Ymax=2

Step 5: Press ENTER to get the graph.

The graph obtained is:

From the graph it can be seen that there is symmetry in the graph.

So, calculate the area of upper portion and then multiply it by two.

So, the shaded section is the twice the region formed from the curve r=3 to r=0.

So, equate the polar equation equal to 0 and get;

r=01+2cosθ=02cosθ=1cosθ=12

For upper region; θ=2π3

And, at r=3 the value of θ is:

3=1+2cosθ2=2cosθ1=cosθ

This gives;

θ=0

The area of the outer loop is twice the area formed by the integration of the polar equation from θ=0 to θ=2π3.

Aouter=2[1202π3[f(θ)]2dθ]=2[1202π3[1+2cosθ]2dθ]=2[1202π3[1+4cos2θ+4cosθ]dθ]

Use the identity 1+cos2θ=2cos2θ,

Aouter=02π3[1+2(1+cos2θ)+4cosθ]dθ=02π3[3+2cos2θ+4cosθ]dθ

Simplify further and get,

Aouter=([3θ+4sinθ+sin2θ]02π3)=[3(2π3)+4sin(2π3)+sin2(2π3)]=[2π+43232]=[2π+332]

Therefore, the value area of the outer loop of r=1+2sinθ is 2π+332.

The inner loop starts and end from the pole r=0.

At r=0:

r=01+2cosθ=02cosθ=1cosθ=12

This gives;

θ=2π3 and θ=4π3

Now, the area of the inner loop is the integration of the polar equation r=1+2sinθ from

θ=2π3 and θ=4π3

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