Double integral help please? polar and cartesian (1 Viewer)

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1. The problem statement, all variables and given/known data

Okay here's the problem:

Consider the region R interior to a circle(of r =2) and exterior to a circle(r=1).

1.Using cartesian coords and double integral, calc the area of annulus.

2. repeat calculation above but using double integral with polar coords


3. The attempt at a solution

So for 1, did I set it up right?

4 * (double integral, both bounds from 0 to 2) of [tex]\sqrt{(4-x^2)}[/tex] dydx - 4 * (double integral, both bounds from 0 to 1) of [tex]\sqrt{1-x^2}[/tex]


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2.

and this is 2, did i set it up right?

First bound is the outer integral, the one for dtheta

(double integral)(first bound 0 to 2 * PI)(Second: -2 to 2) of r dr dtheta - (double integral)(first bound 0 to 2 *PI)(second: -1 to 1) of r dr dtheta

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How do I solve these any further? I completely forgot. Thanks!!
 

tiny-tim

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hi bombz! :smile:

(have a pi: π and a theta: θ and a square-root: √ and an integral: ∫ :wink:)

your cartesian integral is fine (use a trig substitution to solve it), but in your polar integral you can't have negative values of the r coordinate :wink:
 
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Dear tiny tim :)

What would be the lower bound for r then?? Would I just multiply the whole integral by 2 and make the bounds go from 0 to 2 and the other 0 to 1?
 

tiny-tim

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hi bombz! :smile:

you can do this with one double integral instead of two (that's the advantage of polar coordinates in this problem) :wink:

for each value of θ, r goes from 1 to 2, doesn't it?

and then θ goes from 0 to 2π :smile:
 
556
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So for 1, did I set it up right?

4 * (double integral, both bounds from 0 to 2) of [tex]\sqrt{(4-x^2)}[/tex] dydx - 4 * (double integral, both bounds from 0 to 1) of [tex]\sqrt{1-x^2}[/tex]
It's really not clear what you do here.
And if I understand what you did, it's wrong.
The problem did not ask you a short way to calculate the anulus area, but to use a double integral because you might have to exercise on double integrals.

Is this your cartesian integral ?

[tex]\int_{0}^{2}\int_{0}^{2}\sqrt{4-x^2}\ dy dx - 4\int_{0}^{1}\int_{0}^{1}\sqrt{1-x^2}\ dy dx[/tex]
 
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Yes, quinzio, that was my cartesian integral. Except there's a 4 in front of the first integral. You wrote it correctly the way I have it in my work. It did ask me to calculate the area of the annulus though. so can't I do what I did?


And tiny-tim, thank you :) You put me on track with the polar coords :)

so I got the following for polar coordinates then:

(Double integral)(first bound 0 to 2 * PI)(Second bound: 1 to 2) of r dr dtheta

Is that correct?
 
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556
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Polar integral is ok, and very easy of course.

Cartesian integral is not correct.
 
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Polar integral is ok, and very easy of course.

Cartesian integral is not correct.

Why is the Cartesian integral wrong? I take 1/4 slice of the circle, calculate area of the big circle and then calculate the area of the small circle and subtract them out to receive the area of the annulus.
 

tiny-tim

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your cartesian integral is fine :smile:
 

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