Chapter 10.5, Problem 25E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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 = 3 − 6 sin θ

To determine

To calculate: The value of the area between the loops of polar equation r=36sinθ and draw the area by the use of graphing utility.

Explanation

Given:

The polar equation is r=3âˆ’6sinÎ¸.

Formula used:

The area of the polar equation is given by;

A=12âˆ«Î±Î²[f(Î¸)]2dÎ¸

Where, Î±Â andÂ Î² are limits of the integration.

Calculation:

Consider the polar equation r=3âˆ’6sinÎ¸.

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=âˆ’9,Xmax=9,Ymin=âˆ’10Â andÂ Ymax=2

Step 5: Press ENTER to get the graph.

The graph obtained is:

Since, it can be seen that there is symmetry in above graph.

So, the area inside the outer loop is twice the area obtained by integration of the polar equation r=3âˆ’6sinÎ¸ from r=0Â toÂ r=9.

At r=0, the value of Î¸ is;

0=3âˆ’6sinÎ¸3=6sinÎ¸12=sinÎ¸

This gives,

Î¸=Ï€6Â andÂ Î¸=5Ï€6

And, at r=9, the value of Î¸ is;

9=3âˆ’6sinÎ¸6=âˆ’6sinÎ¸âˆ’1=sinÎ¸

This gives;

Î¸=3Ï€2

So, to get the area inside the outer loop, twice the area obtained by integration of the polar equation r=3âˆ’6sinÎ¸ from Î¸=5Ï€6Â toÂ Î¸=3Ï€2.

The area inside the loop is,

Aouter=2[12âˆ«5Ï€/63Ï€/2[f(Î¸)]2dÎ¸]=âˆ«5Ï€/63Ï€/2[3âˆ’6sinÎ¸]2dÎ¸=âˆ«5Ï€/63Ï€/2[9+36sin2Î¸âˆ’36sinÎ¸]dÎ¸

Use the identity 1âˆ’cos2Î¸=2sin2Î¸;

Aouter=âˆ«5Ï€/63Ï€/2[9+18(1âˆ’cos2Î¸)âˆ’36sinÎ¸]dÎ¸=âˆ«5Ï€/63Ï€/2[27âˆ’18cos2Î¸âˆ’36sinÎ¸]dÎ¸=[27Î¸âˆ’9sin2Î¸âˆ’36cosÎ¸]5Ï€/63Ï€/2

Further solve and get,

Aouter=[27Î¸âˆ’9sin2Î¸âˆ’36cosÎ¸

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