What is the Order of Integration in a Double Integral?

[Chenkel]

TL;DR Summary

I'm trying to understand a strange integral in Chris McMullen's book "Essential Calculus-based PHYSICS"

Hello everyone!

I've been reading Mr. McMullen's book and took some curiosity in an equation on the cover art, it is as follows:$$y_{cm} = \frac \rho m \int_{r=0}^R\int_{\theta=0}^\pi (r\sin \theta)rdrd\theta$$I'm trying to understand what it means, firstly the limits of integration for the inner integral are theta, and we're integrating with respect to r; then on the outer integral, the limits of integration are r, and the variable we're integrating with respect to is theta. I'm used to the variable for the limits of integration matching the variable we're integrating with respect to, does this equation make sense to anyone? Let me know what you think, thank you!

 

[kuruman]

It is the integral for finding the center of mass of a semicircular disk of uniform surface mass density $\rho$ using polar coordinates. The definition is $y_{cm}=\frac{1}{m}\int y~dm.$ An area element $dm$ is at $y=r\sin\theta$ and $dm=\rho~dA=\rho(dr)(rd\theta)$. When all this is brought together into a double integral, the limits of integration are over a semicircle of radius $R$.

Look in the book. This might be an example in the center of mass section.

 

[Chenkel]

It is the integral for finding the center of mass of a semicircular disk of uniform surface mass density $\rho$ using polar coordinates. The definition is $y_{cm}=\frac{1}{m}\int y~dm.$ An area element $dm$ is at $y=r\sin\theta$ and $dm=\rho~dA=\rho(dr)(rd\theta)$. When all this is brought together into a double integral, the limits of integration are over a semicircle of radius $R$.

Look in the book. This might be an example in the center of mass section.

I worked out the problem mechanically by hand, however I find I'm still coming to terms with the theory behind the double integral. I found it wasn't possible to come to an answer when the variable in the limits of integration is not the same as the variable we're integrating with respect to, so the equation on the cover art is incorrect if I'm not mistaken. This is my work:$$y_{cm} = \frac \rho m \int_{\theta = 0}^{\pi} \int_{r=0}^R r^2\sin \theta drd\theta$$$$y_{cm} = \frac \rho m \int_{\theta = 0}^{\pi} \frac {r^3} 3 \sin \theta |_{r=0}^Rd\theta$$$$y_{cm} = \frac \rho m (- \frac {R^3} 3 \cos \theta) |_{\theta=0}^\pi$$$$y_{cm} = \frac \rho m \frac {2R^3} 3$$

 

[Delta2]

I worked out the problem mechanically by hand, however I find I'm still coming to terms with the theory behind the double integral. I found it wasn't possible to come to an answer when the variable in the limits of integration is not the same as the variable we're integrating with respect to, so the equation on the cover art is incorrect if I'm not mistaken. This is my work:$$y_{cm} = \frac \rho m \int_{\theta = 0}^{\pi} \int_{r=0}^R r^2\sin \theta drd\theta$$$$y_{cm} = \frac \rho m \int_{\theta = 0}^{\pi} \frac {r^3} 3 \sin \theta |_{r=0}^Rd\theta$$$$y_{cm} = \frac \rho m (- \frac {R^3} 3 \cos \theta) |_{\theta=0}^\pi$$$$y_{cm} = \frac \rho m \frac {2R^3} 3$$

The formula on the cover is correct and you calculated the relevant integral correctly, just replace $$\rho=\frac{m}{\frac{\pi R^2}{2}}$$ and you get the correct result.

 

[Chenkel]

The formula on the cover is correct and you calculated the relevant integral correctly, just replace $$\rho=\frac{m}{\frac{\pi R^2}{2}}$$ and you get the correct result.

That make sense to me. The only part that I might be a little semantic about is if the equation on the cover art is correct, for example, I believe the integral I used is not directly the same as the one on the cover art:

$$\frac \rho m \int_{r=0}^R\int_{\theta=0}^\pi (r\sin \theta)rdrd\theta \neq \frac \rho m \int_{\theta = 0}^{\pi} \int_{r=0}^R r^2\sin \theta drd\theta$$

Notice that one of the integrals has badly defined limits of integration.

 

[Chenkel]

The formula on the cover is correct and you calculated the relevant integral correctly, just replace $$\rho=\frac{m}{\frac{\pi R^2}{2}}$$ and you get the correct result.

I discovered we can also do the approach of distributing dr, so the equation on cover art may be correct:$$\int_{r=0}^R\int_{\theta=0}^\pi (r\sin\theta)rdrd\theta=\int_{r=0}^R(-\cos{\theta}r^2dr)|_{\theta=0}^\pi=\int_{r=0}^R(r^2dr + r^2dr)=\int_{r=0}^R2r^2dr$$
Is this a valid approach?

 

[erobz]

Just a heads up: I think typically you want to flip flop the order of the differentials to work from the inside out.

 

[Delta2]

@Chenkel The order of integration doesn't matter in a double integral as there is Fubini's theorem
https://en.wikipedia.org/wiki/Fubini's_theorem
But yes the order of integration (more specifically, the order at which the differentials are written) matters regarding the way the boundary of integration are written. Yes you are right that the first integral has boundaries of integration 0 to $\pi$ for $r$, and from $0$ to $R$ for $\theta$ which simply isn't correct.