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Finding the Area Enclosed by a Polar Curve

by HolyDesperado
Tags: curve, enclosed, polar
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HolyDesperado
#1
Mar14-09, 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 x-axis.

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CompuChip
#2
Mar14-09, 02:01 PM
Sci Advisor
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P: 4,300
The integration boundaries for theta are given.

[Hint: first find the area between the curve and the x-axis, then find the area under the straight line segment between its ends and subtract appropriately.]
HolyDesperado
#3
Mar14-09, 02:05 PM
P: 16
Quote Quote by CompuChip View Post
The integration boundaries for theta are given.

[Hint: first find the area between the curve and the x-axis, then find the area under the straight line segment between its ends and subtract appropriately.]
are you talking about the inequality from 0 to 1/8? so my boundaries of int. are a=0, b=1/8?

CompuChip
#4
Mar14-09, 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.
HolyDesperado
#5
Mar14-09, 02:38 PM
P: 16
Quote Quote by CompuChip View Post
Yes.
Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates.
integrating from 0 to 1/8 is giving me .355, which isn't right =-\
slider142
#6
Mar14-09, 02:52 PM
P: 898
Why did you include an arclength formula in your OP when the question asks for area?
HolyDesperado
#7
Mar14-09, 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
HolyDesperado
#8
Mar14-09, 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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