- #1
Double integral in polar coordination
- Thread starter rado5
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- Double integral Integral Polar
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Homework Help Overview
The discussion revolves around evaluating a double integral using polar coordinates, specifically addressing the correctness of the integration ranges and the transformation of the area element.
Discussion Character
- Assumption checking, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants explore the transformation of a double integral from Cartesian to polar coordinates, questioning the accuracy of the ranges used for integration. Some express uncertainty about the limits of integration and seek validation of their approach.
Discussion Status
Participants are actively discussing the validity of the integration ranges, with some asserting their correctness based on the relationships in polar coordinates. There is a focus on clarifying misunderstandings regarding the positivity of integrals and the implications for the results obtained.
Contextual Notes
There is an ongoing concern about the ranges of integration, particularly for the variable "r," and how these affect the overall evaluation of the integral. Participants are also referencing specific equations and results from a textbook, indicating a reliance on external sources for validation.
- #2
Susanne217
- 311
- 0
rado5 said:
If you have a double integral on the form I = [tex]\int \int_{R} f(x,y) dy dx[/tex]
Then if you to integrate using polar coordinates then it changes to
I = [tex]\int \int_{R} f(x,y) dA[/tex] where
[tex]dA = r dr d\theta[/tex]
trying adding this and see if you don't get the right result.
- #3
rado5
- 71
- 0
Dear Susanne217,
Thank you very much for your reply, but i have already added [tex]dA = r dr d\theta[/tex]
to get the result. If you calculate it you will see that I had already added it! But my problem is the ranges. I suspect about the ranges.
Thank you very much for your reply, but i have already added [tex]dA = r dr d\theta[/tex]
to get the result. If you calculate it you will see that I had already added it! But my problem is the ranges. I suspect about the ranges.
- #4
Quinzio
- 557
- 1
The quantity you integrate (that is '1') is always positive.
An integral is basically a sum, so a sum of positive terms cannot turn into something negative.
The result of the book [tex](1-\sqrt2 \cong -0.41)[/tex] is negative, so it is wrong.
[tex]=\int_0^{\pi/4} \int_{0}^{\frac{sen\theta}{cos^2\theta}} dr\: d\theta[/tex]
[tex]=\int_0^{\pi/4} {\frac{sen\theta}{cos^2\theta}} \: d\theta[/tex]
[tex]=\left[{\frac{1}{cos\theta}\right]_0^{\pi/4}[/tex]
[tex]=\sqrt2-1[/tex]
Your result is ok.
An integral is basically a sum, so a sum of positive terms cannot turn into something negative.
The result of the book [tex](1-\sqrt2 \cong -0.41)[/tex] is negative, so it is wrong.
[tex]=\int_0^{\pi/4} \int_{0}^{\frac{sen\theta}{cos^2\theta}} dr\: d\theta[/tex]
[tex]=\int_0^{\pi/4} {\frac{sen\theta}{cos^2\theta}} \: d\theta[/tex]
[tex]=\left[{\frac{1}{cos\theta}\right]_0^{\pi/4}[/tex]
[tex]=\sqrt2-1[/tex]
Your result is ok.
Last edited:
- #5
rado5
- 71
- 0
Quinzio said:
Thank you very much for the notification about "positive". But would you please tell me what your idea is about the ranges? especially about the ranges of "r". Are you sure that my ranges are correct?
- #6
Quinzio
- 557
- 1
Yes they are correct because in a polar-cartesian system:
the system
[tex]y = x tg\theta[/tex]
[tex]y = x^2[/tex]
gives [tex]x = tg\theta[/tex]
[tex]r = \sqrt(x^2+y^2)= tg\theta \sqrt(tg^2\theta+1) = tg\theta / cos\theta = sen\theta / cos^2\theta[/tex]
the system
[tex]y = x tg\theta[/tex]
[tex]y = x^2[/tex]
gives [tex]x = tg\theta[/tex]
[tex]r = \sqrt(x^2+y^2)= tg\theta \sqrt(tg^2\theta+1) = tg\theta / cos\theta = sen\theta / cos^2\theta[/tex]
- #7
rado5
- 71
- 0
Dear Quinzio,
Thank you very much for your help.
Thank you very much for your help.
- #8
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