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Finding the Area Enclosed by a Polar Curve 
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#1
Mar1409, 01:57 PM

P: 16

1. The problem statement, all variables and given/known data
Find the area enclosed by the polar curve r = 2 e^(0.9theta) on the interval 0 <= theta <= 1/8 and the straight line segment between its ends. 2. Relevant equations arclength = 3. The attempt at a solution I need help finding the boundaries for this problem. Is it from 0 to pi? I'm not sure how exactly would I go about finding the limits of integration(boundaries) since it starts from the origin and ends at the xaxis. 


#2
Mar1409, 02:01 PM

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P: 4,300

The integration boundaries for theta are given.
[Hint: first find the area between the curve and the xaxis, then find the area under the straight line segment between its ends and subtract appropriately.] 


#3
Mar1409, 02:05 PM

P: 16




#4
Mar1409, 02:29 PM

Sci Advisor
HW Helper
P: 4,300

Finding the Area Enclosed by a Polar Curve
Yes.
Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates. 


#5
Mar1409, 02:38 PM

P: 16




#6
Mar1409, 02:52 PM

P: 898

Why did you include an arclength formula in your OP when the question asks for area?



#7
Mar1409, 03:18 PM

P: 16

i thought it meant arclength, but i guess it is area
so it should be int [a>b] of (1/2(r)^2)dtheta but I still don't know how to set up this problem correctly 


#8
Mar1409, 05:20 PM

P: 16

I figured it out, it's Integral[0>1/8] (1/2*(2 e^(0.9theta))^2)dTheta
thanks everyone 


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