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Trying to find an alternative for solving an integral

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15
1. The problem statement, all variables and given/known data

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2. Relevant equations

Captura de pantalla (524).png


3. The attempt at a solution

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I understand how the integral is solved using cartesian coordinates.

However, I wanted to try to solve it using polar coordinates:

$$\int_0^{\pi/2} cos \theta \sqrt{1+r^2 cos^2 \theta}d \theta\int_{0}^{\sqrt{1-r^2 cos^2 \theta}}r^3dr$$

But it doesn't seem to be a good idea.

Am I wrong or we cannot find a better method than cartesian coordinates for solving this integral?
 

Attachments

Ray Vickson

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1. The problem statement, all variables and given/known data

View attachment 240513

2. Relevant equations

View attachment 240514

3. The attempt at a solution

View attachment 240515

I understand how the integral is solved using cartesian coordinates.

However, I wanted to try to solve it using polar coordinates:

$$\int_0^{\pi/2} cos \theta \sqrt{1+r^2 cos^2 \theta}d \theta\int_{0}^{\sqrt{1-r^2 cos^2 \theta}}r^3dr$$

But it doesn't seem to be a good idea.

Am I wrong or we cannot find a better method than cartesian coordinates for solving this integral?
Your last integral with respect to ##r## also has ##r## in the upper limit; that is meaningless. Not only that, your first integrand (for the ##d \theta## integral) still has an ##r## in it, but that ##r## is nowhere integrated out.
 
207
15
Your last integral with respect to ##r## also has ##r## in the upper limit; that is meaningless. Not only that, your first integrand (for the ##d \theta## integral) still has an ##r## in it, but that ##r## is nowhere integrated out.
Is it meaningless because polar coordinates doesn't apply here or because I made a mistake changing coordinates?
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,710
Is it meaningless because polar coordinates doesn't apply here or because I made a mistake changing coordinates?
Well, let's see: for ##dA = dx \, dy##, ##dA## becomes ##r \, dr \, d \theta## and so
##x \sqrt{1+x^2} \, dA## becomes ## r \cos(\theta) \sqrt{1+r^2 \cos^2(\theta)}\, r \, dr \, d\theta, ## with ##0 \leq r \leq 1## and ##0 \leq \theta \leq \pi/4.## You could do the ##r##-integral (with ##\theta## fixed) to get a function of ##\theta## that would then need to be integrated over ##\theta## from 0 to ##\pi/4##; or, you could do the ##\theta##-integral (with fixed ##r##) to get a function of ##r## that would need integrating from 0 to 1. In either case you would not get what you wrote.
 
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