The problem is to find the area of the region enclosed by the loop of the strophoid r=2cos(θ)−sec(θ). I am having trouble figuring out how to get the limits of integration for this one. Some helpful methods/tips on how to go through the process would be appreciated.

Working on half of the surface (from θ = 0 to θ = pi/4) :
An infinitesimal shape to compute the surface would be a right triangle, of leg r(θ) and height r(θ).dθ, hence of area r²(θ).dθ/2

So, S = 2. ∫ r²(θ).dθ/2, from 0 to pi/4.
S = ∫ r²(θ).dθ = 2 - pi/2

if I do no mistake.
I guess the estimation of that integral should not be hard.

Thanks, if I integrated correctly, the answer should be .4292

Yes, up to 4 decimal places.

Do you have any idea what the limits would be for this one? Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside the curve r=2. I thought it would be where cos(2θ)=1/2 which means 2θ=pi/3 and 5pi/3, so θ = pi/6 and 5pi/6, but that gives me a wrong answer.

Then consider only θ from 0 to pi/6. You'll have to multiply by 4 the following integral :

∫ r²(θ).dθ/2 - pi/3

where pi/3 is the area of the portion of the disk lying between θ = 0 and θ = pi/6.

∫ r²(θ).dθ/2 = ∫ 8.cos(2θ).dθ/2 = 2.sin(pi/3) = √3

...

And finally, S = 4√3 - 4.pi/3 ~ 2.7394

geez makes me feel like my calc 3 class is bad >_>'

we are going over vectors, space geometry and vector value functions right now