# Finding the Area Enclosed by a Polar Curve

Tags: curve, enclosed, polar
 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.
 Sci Advisor HW Helper 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.]
P: 16
 Quote by CompuChip 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?

 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.
P: 16
 Quote by CompuChip 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 =-\
 P: 898 Why did you include an arclength formula in your OP when the question asks for area?
 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
 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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