Finding the Area Enclosed by a Polar Curve

[HolyDesperado]

Homework Statement

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.

Homework Equations

arclength =

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

 

[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.]

 

[HolyDesperado]

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]

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]

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]

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

 

[HolyDesperado]

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

 

[HolyDesperado]

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

thanks everyone