   Chapter 10.5, Problem 23E

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 = 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=36sinθ.

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=36sinθ.

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=9,Xmax=9,Ymin=10 and 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=36sinθ from r=0 to r=9.

At r=0, the value of θ is;

0=36sinθ3=6sinθ12=sinθ

This gives,

θ=π6 and θ=5π6

And, at r=9, the value of θ is;

9=36sinθ6=6sinθ1=sinθ

This gives;

θ=3π2

So, to get the area inside the outer loop, twice the area obtained by integration of the polar equation r=36sinθ from θ=5π6 to θ=3π2.

The area inside the loop is,

Aouter=2[125π/63π/2[f(θ)]2dθ]=5π/63π/2[36sinθ]2dθ=5π/63π/2[9+36sin2θ36sinθ]dθ

Use the identity 1cos2θ=2sin2θ;

Aouter=5π/63π/2[9+18(1cos2θ)36sinθ]dθ=5π/63π/2[2718cos2θ36sinθ]dθ=[27θ9sin2θ36cosθ]5π/63π/2

Further solve and get,

Aouter=[27θ9sin2θ36cosθ

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