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- Thread starter SMA_01
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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.

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.

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- #5

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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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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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