Having trouble with double integrals in polar coordinates?

Maybe if you gave the equation you'd get something more definite. But for polar coordinates what i'd do personally is convert it to a set of parametric equations and integrate those separately with respect to the parameter.

R is usually given by a polar equation, i'm not quite sure what you're asking.

 

again it would be helpful if you gave the context. the radius is just the distance from the origin. where do the double integrals fit in?

 

sorry, just to clarify.. [itex] r = \frac{1}{9 \cos{t}}[/itex] for [itex]-\pi \le t \le \frac{\pi}{2}[/itex] and [itex] r = 1 [/itex] right?

For the lower bound on r, I believe it is [itex]\frac{-1}{9} [/itex], but i'm not entirely sure about that, maybe someone else can look at this.

In terms of the iterated integral here's a few tips:
[tex] x = r \cos{t} [/tex]
[tex] y = r \sin{t} [/tex]
[tex] dA = r dr dt [/tex]

and you're integrating a function of the form [itex] f(x, y) [/itex].

 

would you mind? can you please ans this.... please find the polar form of 1/4i also the 1/z? please