Double integral with polar coordinates

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Amaelle
Messages
318
Reaction score
55
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
 
@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...
 
Reply
  • Like
Likes   Reactions: SammyS