Question

Find the area of the region that lies inside the first curve and outside the second curve

Answer 1

\[r=3\cos \theta~~~~r = 1+\cos \theta \]

Diagram: @ganeshie8

When you have problems as such does the basic execution of integrals still apply, such as f(x)-g(x) upper - lower?

Answer 2

\[A=2 \int\limits_{0}^{\pi/3} 1/2 [(3\cos \theta)^2-(1+\cos \theta)^2] d \theta\] like this would be your integral right? Is it always like this?

Answer 3

the key thing is to sketch the are we seek and finding the intersection points

Answer 4

setting up integral becomes easy and makes more sense after we get some hold of how the region looks like

Answer 5

Yeah I got that part, just trying to understand if it's always similar as such, also r = r to find theta but I did x2 since it was symmetrical otherwise I'd have to do pi/3 to 5pi/3 as well right?

But mainly my question is, are these questions always like this or are there problems where you'd add the x and y?

Answer 6

I should be asking why \[A=2 \int\limits\limits_{0}^{\pi/3} 1/2 [(3\cos \theta)^2-(1+\cos \theta)^2] d \theta\] is not \[A=2 \int\limits\limits_{0}^{\pi/3} 1/2 [(3\cos \theta)^2+(1+\cos \theta)^2] d \theta\] sorry if this is confusing..

Answer 7

I'll answer that question on why we're subtracting, but let me point out that integral is wrong

Answer 8

your integral gives below area, right ?