- #1
Double integral in polar coordination
- Thread starter rado5
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- Double integral Integral Polar
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SUMMARY
The discussion centers on evaluating a double integral using polar coordinates, specifically addressing the integral I = ∫∫_R f(x,y) dA, where dA = r dr dθ. Participants confirm the correctness of the integration process and emphasize the importance of proper range selection for the variable r. The conclusion drawn is that the ranges provided in the original problem are accurate, as they align with the polar-coordinate transformations of the Cartesian equations y = x tan(θ) and y = x². The negative result from the textbook is deemed incorrect due to the nature of integrals summing positive quantities.
PREREQUISITES- Understanding of double integrals in calculus
- Familiarity with polar coordinates and their transformations
- Knowledge of trigonometric functions and identities
- Experience with evaluating integrals in multiple dimensions
- Study the conversion of Cartesian coordinates to polar coordinates in integrals
- Learn about the properties of definite integrals and their ranges
- Explore the implications of integrating non-negative functions
- Review examples of double integrals in polar coordinates for better comprehension
Students and educators in calculus, mathematicians working with integrals, and anyone seeking to deepen their understanding of polar coordinate integration techniques.
- #2
Susanne217
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rado5 said:
If you have a double integral on the form I = \int \int_{R} f(x,y) dy dx
Then if you to integrate using polar coordinates then it changes to
I = \int \int_{R} f(x,y) dA where
dA = r dr d\theta
trying adding this and see if you don't get the right result.
- #3
rado5
- 71
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Dear Susanne217,
Thank you very much for your reply, but i have already added dA = r dr d\theta
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 dA = r dr d\theta
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
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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 (1-\sqrt2 \cong -0.41) is negative, so it is wrong.
=\int_0^{\pi/4} \int_{0}^{\frac{sen\theta}{cos^2\theta}} dr\: d\theta
=\int_0^{\pi/4} {\frac{sen\theta}{cos^2\theta}} \: d\theta
=\left[{\frac{1}{cos\theta}\right]_0^{\pi/4}
=\sqrt2-1
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 (1-\sqrt2 \cong -0.41) is negative, so it is wrong.
=\int_0^{\pi/4} \int_{0}^{\frac{sen\theta}{cos^2\theta}} dr\: d\theta
=\int_0^{\pi/4} {\frac{sen\theta}{cos^2\theta}} \: d\theta
=\left[{\frac{1}{cos\theta}\right]_0^{\pi/4}
=\sqrt2-1
Your result is ok.
Last edited:
- #5
rado5
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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
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Yes they are correct because in a polar-cartesian system:
the system
y = x tg\theta
y = x^2
gives x = tg\theta
r = \sqrt(x^2+y^2)= tg\theta \sqrt(tg^2\theta+1) = tg\theta / cos\theta = sen\theta / cos^2\theta
the system
y = x tg\theta
y = x^2
gives x = tg\theta
r = \sqrt(x^2+y^2)= tg\theta \sqrt(tg^2\theta+1) = tg\theta / cos\theta = sen\theta / cos^2\theta
- #7
rado5
- 71
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Dear Quinzio,
Thank you very much for your help.
Thank you very much for your help.
- #8
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