# Using the Product of the Integrals is integrals of Product Magic in multivar

Using the "Product of the Integrals is integrals of Product" Magic in multivar

This isn't actually a question on a homework problem, but It is coursework related so I thought it belonged here. I've noticed a lot of magic in Griffiths and other textbooks, where say, in integrating in spherical co-ordinates, you simply do the theta, phi, and r integrals, and them multiply them all together...

but I thought this was not a valid way to do integration... like it makes sense to me, but I can't justify to myself why it would always be true that one could do this... is it maybe only true when the function your integrating doesn't depend on all three variables, or you can get it into a form where the product of these things don't?

For instance, Int (e^-x^2-y^2)dxdy.... across all of R^2 this integral of this is equal to the integral of e^-x^2 dx times from 0-> infinity the integral of e^-y^2 dy multiplied together from 0 to infinity... Why? and why is it legit to do this "pulling out dphi and multiplying by that integral" like Griffiths does in his electrodynamics?

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Cyosis
Homework Helper

$$\int \int e^{-x^2-y^2}dxdy=\int \int e^{-x^2}e^{-y^2}dxdy=\int e^{-y^2} \left(\int e^{-x^2}dx \right)dy=\int e^{-y^2} dy \int e^{-x^2} dx$$

Step 2 to 3: $e^{-y^2}$ is a constant with respect to x, so you can take it in front of the integral.
step 3 to 4: $\int e^{-x^2}dx$ is a constant with respect to dy, so you can take it in front of the integral.

In Griffiths you're probably talking about integrals like:

$$W=\frac{\epsilon_0}{2}\int_{all space}E^2 dV=\int_0^R \int_0^\pi \int_0^{2\pi} E^2 r^2 \sin \theta d\phi d\theta dr= \int_0^R E^2 r^2 dr \int_0^\pi \sin \theta d\theta \int_0^{2\pi} d\phi$$

This is because, E usually has a dependence on r alone, had it been $\vec{E(r,\theta,\phi)}$ this would only have worked if $\vec{E(r,\theta,\phi)}=E(r)E(\theta)E(\phi)$

For example:
$$\int \int 2 \cos (xy) \sin (xy)\, dxdy \neq 2 \int \cos (xy)\, dx \int \sin (xy)\, dy$$

Try it out.

Last edited:
HallsofIvy
Homework Helper

If the limits of integration are constant and the function is "separable", f(x,y)= g(x)h(y), that is valid.

$$\int_{x= a}^{b}\int_{y= c}^d g(x)h(y) dydx= \int_{x=a}^f g(x)\left(\int_{y= c}^d h(y)dy\right) dx$$
Since h is a function of y only and the limits of integration, c and d, are constant, the integration inside the parentheses is a constant and so can be taken out of the second integral:
$$\left(\int_{y= c}^d h(y)dy\right)\left(\int_{x= a}^b g(x)dx\right)$$

Great; Thanks guys!