# Trying to find an alternative for solving an integral

#### JD_PM

1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 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?

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#### Ray Vickson

Science Advisor
Homework Helper
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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.

• JD_PM

#### JD_PM

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
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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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• JD_PM

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