Having trouble with double integrals in polar coordinates?

  • Thread starter SMA_01
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  • #1
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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?
 

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  • #2
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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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  • #3
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@gordonj005: I mean I'm confused as to what steps to take to find the radius, r.
 
  • #4
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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
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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
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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].
 
  • #7
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would you mind? can you please ans this.... please find the polar form of 1/4i also the 1/z? please
 

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