Double integral/Polar coordinates example

[iocal]

Hi guys,

I am looking for help on a particular case which I can solve in cartesian coordinates but not in polar ones. In principle I should be able to do so in both but there is something that I am missing, it seems. The example is:

Calculate the area of x^2 + y^2 for -1<x<1 and -1<y<1.

The result, easily found in cartesian coordinates is 8/3 but in polar coordinates I do not know how to set the borders for θ correctly so as to get that result.
Any help greately appreciated, thank you!

 

[Deleted member 4993]

Hi guys,

I am looking for help on a particular case which I can solve in cartesian coordinates but not in polar ones. In principle I should be able to do so in both but there is something that I am missing, it seems. The example is:

Calculate the area of x^2 + y^2 for -1<x<1 and -1<y<1.

The result, easily found in cartesian coordinates is 8/3 but in polar coordinates I do not know how to set the borders for θ correctly so as to get that result.
Any help greately appreciated, thank you!

I am on vacation - and I do not have proper reference books - so I have not actually tried to solve the problem, but I'll give my thoughts.

I'll try the limits of Θ to be 0 to π/4 and multiply the result by 8.

I would have to first integrate wrt r.

The limits of r would be 0 to sec(Θ).

 

[DrPhil]

Hi guys,

I am looking for help on a particular case which I can solve in cartesian coordinates but not in polar ones. In principle I should be able to do so in both but there is something that I am missing, it seems. The example is:

Calculate the area of x^2 + y^2 for -1<x<1 and -1<y<1.

The result, easily found in cartesian coordinates is 8/3 but in polar coordinates I do not know how to set the borders for θ correctly so as to get that result.
Any help greately appreciated, thank you!

I'll try the limits of Θ to be 0 to π/4 and multiply the result by 8.

I would have to first integrate wrt r.

The limits of r would be 0 to sec(Θ).

I want to write that out in LaTeX.
Remembering that the increment of area in polar coordinates is \(\displaystyle (r\ d\theta)dr\),
and replacing \(\displaystyle x^2 + y^2 \to r^2\) for the integrand,

\(\displaystyle \displaystyle 8 \int_0^{\pi /4}\left[\int_0^{\sec \theta} r^2\ r\ dr\right] d\theta \)

I worked this through, and came up with the answer 8/3 . It was a lot easier in Cartesian coordinates!

 

[iocal]

I want to write that out in LaTeX.
Remembering that the increment of area in polar coordinates is \(\displaystyle (r\ d\theta)dr\),
and replacing \(\displaystyle x^2 + y^2 \to r^2\) for the integrand,

\(\displaystyle \displaystyle 8 \int_0^{\pi /4}\left[\int_0^{\sec \theta} r^2\ r\ dr\right] d\theta \)

I worked this through, and came up with the answer 8/3 . It was a lot easier in Cartesian coordinates!

Thanks a bunch! But can you please tell me why you are multiplying with 8 and also say a few words about the new limits of the integrals?

 

[HallsofIvy]

Thanks a bunch! But can you please tell me why you are multiplying with 8 and also say a few words about the new limits of the integrals?

Because in taking \(\displaystyle \theta\) from 0 to \(\displaystyle \pi/2\) you are staying in the first quadrant while taking r from 0 to \(\displaystyle sec(\theta)\) you are restricting y< x. That is you are actually integrating over 1/8 of the entire plane and then using "symmetry" to argue that the true integral is 8 times that value.

 

[iocal]

Because in taking \(\displaystyle \theta\) from 0 to [itex]\pi/2[/itex] you are staying in the first quadrant while taking r from 0 to \(\displaystyle sec(\theta)\) you are restricting y< x. That is you are actually integrating over 1/8 of the entire plane and then using "symmetry" to argue that the true integral is 8 times that value.

Okay good but I think it will become clearer once I study it further, do you happen to know a good link for that?
Thanks.