Double integral with polar coordinates

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SUMMARY

The forum discussion centers on evaluating a double integral using polar coordinates. The user initially applies the integral I = \int_0^{\pi/4} \int_0^{\sqrt{2+2\cos{2\theta}}} 3r^2 dr d\theta and simplifies it to 8 \int_0^{\pi/4} \cos^4{\theta} d\theta. A mistake occurs during the integration process, where the user mistakenly integrates 8 \cos^5{\theta} instead of 8 \cos^4{\theta}. The discussion highlights the importance of careful calculation in integral evaluation and LaTeX formatting.

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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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