Double integral help please? polar and cartesian

[bombz]

Homework Statement

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

The Attempt at a Solution

So for 1, did I set it up right?

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


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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]

hi bombz!

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

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

 

[bombz]

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]

hi bombz!

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

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

and then θ goes from 0 to 2π

 

[Quinzio]

So for 1, did I set it up right?

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

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 ?

$$\int_{0}^{2}\int_{0}^{2}\sqrt{4-x^2}\ dy dx - 4\int_{0}^{1}\int_{0}^{1}\sqrt{1-x^2}\ dy dx$$

 

[bombz]

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?

 

[Quinzio]

Polar integral is ok, and very easy of course.

Cartesian integral is not correct.

 

[bombz]

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]

your cartesian integral is fine