# Finding the Area Enclosed by a Polar Curve

1. Mar 14, 2009

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.

http://img5.imageshack.us/img5/7910/whatareloi.jpg [Broken]

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Last edited by a moderator: May 4, 2017
2. Mar 14, 2009

### 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.]

3. Mar 14, 2009

are you talking about the inequality from 0 to 1/8? so my boundaries of int. are a=0, b=1/8?

4. Mar 14, 2009

### 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.

5. Mar 14, 2009

integrating from 0 to 1/8 is giving me .355, which isn't right =-\

6. Mar 14, 2009

### slider142

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

7. Mar 14, 2009

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. Mar 14, 2009