Homework Help: Having trouble with double integrals in polar coordinates?

1. Nov 17, 2011

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?

2. Nov 17, 2011

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.

Last edited: Nov 17, 2011
3. Nov 17, 2011

SMA_01

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

4. Nov 17, 2011

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?

5. Nov 17, 2011

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

6. Nov 17, 2011

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 $f(x, y)$.

7. Nov 23, 2011

mccoy9704

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