Double integral with polar coordinates

Click For Summary
The discussion revolves around solving a double integral using polar coordinates, where the original poster presents their method and expresses confusion over an incorrect result. They initially apply a transformation to the integral but struggle with the integration of cos^4. A participant points out that the original poster mistakenly integrated 8*cos^5 instead of 8*cos^4, leading to the error. Additionally, there are comments about LaTeX formatting issues that affected readability. The conversation concludes with the original poster acknowledging the mistake and thanking the contributors for their help.
Amaelle
Messages
309
Reaction score
54
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!
 
Physics news on Phys.org
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:
 
@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...
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...