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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

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BuyFindarrow_forward

Calculus

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

Finding the Area of a Polar Region In Exercises 103-108, use a graphing utility to graph the polar equation. Find the area of the given region analytically.

Common interior of r = 5 2 sin θ and r = 5 + 2 sin θ

To determine

To calculate: The area of the region analytically. Also graph common interior of r=52sinθ and r=5+2sinθ using a graphing utility.

Explanation

Given:

The graph of the function is r=52sinθ and r=5+2sinθ

Calculation:

Consider the equation, r=52sinθ and r=5+2sinθ

Let’s first draw the graph.

To graph polar equation, use Ti83 calculator. The following steps are needed.

Step 1. Press on button.

Step 2. Press MODE button and choose ‘funct” and in funct choose Pol.

Step 3. prees Y= button and write r1=52sinθ and r2=5+2sinθ

Step 4. Press GRAPH button.

And the graph of the polar equation is obtained as below:

The area of the region between the circle 52sinθ and 5+2sinθ and the radical line θ=0 and θ=π is

A1=120π(5+2sinθ)2dθ

That is,

A1=12[0π(2520sinθ+4sin2θ)dθ ]

That is,

A1=12[250πdθ 200πsinθdθ +40πsin2θdθ ]

Use the identity: sin2θ=1cos2θ2

Thus,

A1=12[250πdθ 200πsinθdθ +40π(1cos2θ2)dθ ]=12[250πdθ 200πsinθdθ +20πdθ 20π(cos2θ)dθ ]=12[25[θ]0π20[cosθ]0π+2[θ]0π2[sin2θ2]0π]

Apply the limits with the help of the property:

[F(x)]ab=F(b)F(a)

Then,

A1=12[25[π0]+20[cos(π)cos(0)]+2[π0][sin2(π)sin2(0)]]=12[25π+20[(1)(1)]+2π[00]]=12

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