Find Area with Polar Coordinates: Help for Exams

[Tyzall]

Hello, I actually have an exam coming on series, sequences, polar coordinates and parametric equations. The only major thing I'm having problems with is finding area with polar coordinates. Especially when it comes to finding the "area under both curves" for example:"Find the area under both curves: r = cos(x) r = sin(x)" If anyone is proficient in this material and understands this (I'm sure there are a numerous amount of you on here) any help would be appreciated. Thanks!

 

[matiasmorant]

the area in polar coordinates is $$\int \int rdrd\theta$$ setting the limits of integration so they describe the region we are considering. let's solve your example.
I suggest you visualize the region we are considering. r= sin(q) is a circle of radius 1 centered at (0,1/2) and r= cos(q) is a circle centered at (1/2,0). you mightnot be sure that they are are circles (which they are) but you don't need to know that. by sketching the functions, you will see that they are some sort of intersecting circular areas, which is all you need to know.
know let's find the limits to describe the region of intersection. the region is in the first cuadrant so 0<q<Pi/2, since there is a value of r in this region for all the values of q mentioned. however, 0<r<Sin(q) when 0<q<Pi/4 and 0<r<Cos(q) when Pi/4<q<Pi/2. this wolud be a geometrical description of the region and there is no need to find out the algebraic interplay of the equations, although we could have done it that way too. a mental sketch is unavoidably needed.
so, above we foud the limits. so the desired integral is
$$\int _0^{\text{Pi}/2}\int _0^{\text{Sin}[\theta ]}rdrd\theta +\int _{\text{Pi}/2}^{\text{Pi}/4}\int _0^{\text{Cos}[\theta ]}rdrd\theta$$

now it's just a matter of solving the integral. you should get $$\frac{1}{16} (2+\pi )$$

the hardest and most insightful part is setting up the limits. this is all that gives you a true understanding of the problem.