   Chapter 10.5, Problem 22E

Chapter
Section
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 = 2 ( 1 + 2 sin θ )

To determine

To calculate: The value of the area between the loops of polar equation r=2(1+2sinθ) and to draw the area by the use of graphing utility.

Explanation

Given:

The polar equation r=2(1+2sinθ).

Formula used:

The area of the polar equation is:

A=12αβ[f(θ)]2dθ

Where, α and β are limits of the integration.

Calculation:

Consider the polar equation r=2(1+2sinθ).

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=4,Xmax=4,Ymin=1 and Ymax=6

Step 5: Press ENTER to get the graph.

The graph obtained is:

Area of shaded region=Area of outer regionarea of inner region=AouterAinner

For θ=π2,

r=2(1+2sin(π2))=2(3)=6

And, the area of outer region is bounded by r=0 to r=6.

The value of θ at r=0 is,

0=2(1+2sinθ)0=1+2sinθ12=sinθ

This gives;

θ=π6 and θ=7π6

The area of the outer region bounded by θ=π6 to θ=π2 and since the above graph has symmetry.

So, area of the outer region is;

Ainner=2[12π/6π/2[f(θ)]2dθ]=π/6π/2[2(1+2sinθ)]2dθ=π/6π/24(1+4sinθ+4sin2θ)dθ

Use the identity, 1cos2θ=2sin2θ,

A=π/6π/24(1+4sinθ+2(1cos2θ))dθ=4π/6π/2(1+4sinθ+22cos2θ)dθ=4π/6π/2(3+4sinθ2cos2θ)dθ

Further solve and get,

A=4[3θ4cosθsin2θ]π/6π/2=4[(3(π2

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