Having trouble with double integrals in polar coordinates?

[SMA_01]

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?

 

[gordonj005]

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.

 

[SMA_01]

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

 

[gordonj005]

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?

 

[SMA_01]

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

 

[gordonj005]

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).

 

[mccoy9704]

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