Double integral with polar coordinates

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Homework Help Overview

The discussion revolves around evaluating a double integral using polar coordinates. Participants explore different methods for integration and express confusion regarding the correctness of their approaches.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts an alternative method for solving a double integral but encounters issues with the final result. Other participants provide a detailed breakdown of a method involving integration by parts and the use of trigonometric identities.

Discussion Status

Some participants have offered guidance on the integration process, while others have pointed out potential errors in calculations. There is acknowledgment of the original poster's correct approach, but confusion remains regarding specific steps in the integration.

Contextual Notes

Participants note issues with LaTeX formatting in the discussion, which may have affected the clarity of shared mathematical expressions.

Amaelle
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Homework Statement
look at the image
Relevant Equations
polar coordinates
Greetings!
I have the following integral
1630000459327.png

and here is the solution of the book (which I understand perfectly)
1630000520759.png

1630000549035.png


I have an altenative method I want to apply that does not seems to gives me the final resultMy method
1630002484185.png

which doesn't give me the final results!
where is my mistake?
thank you!
 
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You can write \begin{align*}
I &= \int_0^{\pi/4} \int_0^{\sqrt{2+2\cos{2\theta}}} 3r^2 dr d\theta \\
&= \int_0^{\pi/4} \cos{\theta} \left( 2+2\cos{2\theta}\right)^{3/2} d\theta \\
&= 8 \int_0^{\pi/4} \cos^4{\theta} d\theta
\end{align*}then\begin{align*}
\int \cos^4{x} dx &=\int \dfrac{(e^{ix} + e^{-ix})^4}{16} dx \\
&= \int \left(\dfrac{e^{4ix} + e^{-4ix} + 4e^{2ix} + 4e^{-2ix} + 6}{16} \right) dx \\
&= \int \left( \dfrac{1}{8} \cos{4x} + \dfrac{1}{2} \cos{2x} + \dfrac{3}{8} \right) dx \\
&= \dfrac{1}{32} \sin{4x} + \dfrac{1}{4} \sin{2x} + \dfrac{3}{8} x \bigg{|}_{0}^{\pi/4} \\
&= \dfrac{1}{4} + \dfrac{3\pi}{32}
\end{align*}
 
Last edited:
ergospherical said:
You can write
\begin{align*}
I &= \int_0^{\pi/4} \int_0^{\sqrt{2+2\cos{2\theta}}} 3r^2 dr d\theta \\
&= \int_0^{\pi/4} \cos{\theta} \left( 2+2\cos{2\theta}\right)^{3/2} d\theta \\
&= 8 \int_0^{\pi/4} \cos^4{\theta} d\theta
\end{align*}
then
\begin{align*}
\int \cos^4{x} dx &=\int \dfrac{(e^{ix} + e^{-ix})^4}{16} dx \\
&= \int \left(\dfrac{e^{4ix} + e^{-4ix} + 4e^{2ix} + 4e^{-2ix} + 6}{16} \right) dx \\
&= \int \left( \dfrac{1}{8} \cos{4x} + \dfrac{1}{2} \cos{2x} + \dfrac{3}{8} \right) dx \\
&= \dfrac{1}{32} \sin{4x} + \dfrac{1}{4} \sin{2x} + \dfrac{3}{8} x \bigg{|}_{0}^{\pi/4} \\
&= \dfrac{1}{4} + \dfrac{3\pi}{32}
\end{align*}
thank you but your code is not readable
 
Amaelle said:
thank you but your code is not readable
Yeah I don't know why that's happening, the LaTeX is fine and works when I run it on Overleaf. :frown:
 
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@ergospherical thank you very much!
indeed my approach was correct, I only messed up with the calculation when I was trying to integrate 8*cos^5 instead of 8*cos^4
thank you again!
 
ergospherical said:
Yeah I don't know why that's happening, the LaTeX is fine and works when I run it on Overleaf
It looks like you had no space before \begin and left off the ## delimiters at start/end. I fixed it but you deleted it as I was fixing it...
 
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