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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 103-108, use a graphing utility to graph the polar equation. Find the area of the given region analytically.

Between the loops of r = 4 + 8 sin θ

To determine

To graph: The polar equation, r=2+4sinθ and evaluate the area of the region between the loops.

Explanation

Given:

The polar equation is r=2+4sinθ.

Graph:

Consider the polar equation is r=2+4sinθ.

Follow the steps of ti-83 to obtain graph of the equation.

Step 1: In the TI-83 graphical calculator, press the MODE button.

Step 2: Press the down arrow key twice to “Radian” and press ENTER.

Step 3: Press down the arrow key to “Func”. Then press right key to “Pol” and press ENTER again.

Step 4: Press CLEAR.

Step 5: Press “Y=”.

Step 6: Enter the equation 2+4sinθ.

Step 7: press window button and then set up window as Xmin=6,Xmax=6,Ymin=1 and Ymax=7

Step 8: Press GRAPH.

Step 9: Press ZOOM.

Step 10: Scroll down to 0: Zoom Fit and press ENTER.

For area of the inner loop of the curve r=2+4sinθ, equate the curve equal to zero as the inner loop touches the origin. So as to find the angles or the limit for the integral.

2+4sinθ=04sinθ=2sinθ=24sinθ=12

Since, the sine of the angle is negative in 3rd and 4th quadrant.

The value θ for which sinθ=12 in 3rd quadrant is π+π6=7π6.

The value θ for which sinθ=12 in 4th quadrant is 2ππ6=11π6.

Therefore, the interval for the area of inner loop is [7π6,11π6].

The area for a continuous polar curve r=f(θ) between the two radial lines θ=α and θ=β which is non negative on the interval [α,β] where 0<βα<2π is,

A=12αβ[f(θ)]2dθA=12αβr2dθ

Consider the equation for the polar curve r=2+4sinθ.

Since the interval for the area for one petal lies between the interval [7π6,11π6].

Substitute 7π6 for α, 11π6 for β and 2+4sinθ for r.

A=127π611π6(2+4sinθ)2dθ=127π611π6(4+16sin2θ+16sinθ)dθ

Apply the power reducing formula, cos2θ=12sin2θ.

A=127π611π6(4+16(1cos2θ2)+16sinθ)dθA=127π611π6(4+8(1cos2θ)+16sinθ)dθ=127π611π6(4+88cos2θ+16sinθ)dθ=127π611π6(128cos2θ+16sinθ)dθ

Integrate with respect to θ and apply the limits,

A=12[12θ8sin2θ216cosθ]7π611π6A=12[(12(11π6)8sin2(11π6)216cos(11π6))(12(7π6)8sin2(7π6)216cos(7π6))]=12[(22π4sin(11π3)16cos(11π6))(14π4sin(7π3)16cos(7π6))]=8π+1232

Simplify,

A=4π63

Thus, the area of the inner loop of r=2+4sinθ is 4π63 units.

For area of the outer loop of the curve r=2+4sinθ, equate the curve equal to zero

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