Finding the Area Enclosed by a Polar Curve

HolyDesperado

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.

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

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

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)dtheta


but 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