# Having trouble with double integrals in polar coordinates?

I'm having trouble figuring out how to find what "r" is. I know r is the radius, but how do I go about finding it? Like what do I look for in a particular problem?

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.

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@gordonj005: I mean I'm confused as to what steps to take to find the radius, r.

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?

Okay here is a problem from my homework, and I'm stuck on finding the lower bound for r:

Graph r=1/(9cos(t)) for -∏/2≤t≤∏/2 and r=1. Then write an iterated integral in polar coordinates representing the area inside the curve r=1 and to the right of r=1/(9cost) . (Use t for theta in your work.)

I got the bounds for t -arccos(1/9) ≤t≤ arccos(1/9) and I have the upper bound for r to be 1, but I can't get the lower bound, any ideas?

Thank you

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

For the lower bound on r, I believe it is $\frac{-1}{9}$, 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:
$$x = r \cos{t}$$
$$y = r \sin{t}$$
$$dA = r dr dt$$

and you're integrating a function of the form $f(x, y)$.

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