# Please clear my mathematical physics confusion regarding integration.

EDIT: If multiple Riemann integral is not ok, then please consider Lebesgue integral.

Consider Cartesian coordinate system:

Let there be a cubic charge $$V'$$ of side $$a$$ units with uniform charge density $$\rho'$$ with its center lying at the center of the Cartesian coordinate system.

Let there be a square Gaussian surface $$S$$ parallel to $$y$$-$$z$$ plane with side '$$2a$$' units whose center lies in $$x$$-axis '$$a$$' units away from origin. The Cartesian coordinates of four vertices of the square Gaussian surface are as shown in the figure below:

EDIT:

It may seem from the diagram that a portion of the Gaussian surface coincides with one of the sides of cube. But that is wrong. Indeed $$V' \cap S=\phi$$. The Gaussian surface plane is at $$x=a$$ units while the two sides of the cube parallel to $$y$$-$$z$$ plane is at $$x=a/2$$ and $$x=-a/2$$ units.

Let the Cartesian coordinate of any point on the cubic charge be $$(x',y',z')$$

Let the Cartesian coordinate of any point on the Gaussian surface be $$(a,y,z)$$

Let $$\hat{n}$$ be unit area vector at a point on the Gaussian surface.

Let the Coulomb's constant be $$k$$

We need to find the electric flux through the Gaussian surface. The direct approach is:

$$k\ \rho' \iint_S \big[\iiint_{V'} \dfrac{(a-x')\hat{i} + (y-y')\hat{j} + (z-z')\hat{k}}{[(a-x')^2 + (y-y')^2 + (z-z')^2]^{3/2}} dx'dy'dz' \big] \cdot (\hat{n})\ dy dz$$

$$=k\ \rho' \iint_S \big[\iiint_{V'} \dfrac{(a-x')\hat{i} + (y-y')\hat{j} + (z-z')\hat{k}}{[(a-x')^2 + (y-y')^2 + (z-z')^2] ^{3/2}} \cdot (\hat{i})\ dx'dy'dz' \big] dy dz$$

$$=k\ \rho' \int^a_{-a} \int^a_{-a} \big[\int^{a/2}_{-a/2} \int^{a/2}_{-a/2} \int^{a/2}_{-a/2} \dfrac{a-x'}{[(a-x')^2 + (y-y')^2 + (z-z')^2]^{3/2}} dx'dy'dz' \big] dy\ dz$$

Question 1: Did I apply the order of integration correctly?

I am confused because our domain of integration is $$V' \times S$$. So from a five dimensional view, I do not think the domain of integration be necessarily a $$5$$-$$D$$ rectangle. Please clear my confusion.

• I have edited the question and also corrected the integrand expression. Please have a look at it and please try to answer. Commented Dec 11, 2022 at 5:24
• OK. I understand now that the Gaussian surface and the cube don't touch. Then there should be no convergence problem with the integrals. In fact, in the five dimensional region of integration your integrand is bounded. Then the order of integration does not matter (Fubini). The $x$ in your integrand should be the constant $a$. Commented Dec 11, 2022 at 7:28
• Yes, $x$ is to be replaced with $a$. But are the limits of integration correct? Commented Dec 11, 2022 at 7:30
• Why is the domain $V' \times S$ a five dimensional rectangle? Commented Dec 11, 2022 at 7:31
• This domain is without doubt $[-a,a]\times[-a,a]\times[-a/2,a/2]\times[-a/2,a/2]\times[-a/2,a/2]\,.$ I am counting five dimensions. It is without doubt a rectangle. Please don't ask me about naming conventions in mathematics. Commented Dec 11, 2022 at 8:44

The $$x$$ should be an $$a$$.
What I find more confusing is that the electric field of the cube should be $$\mathbf E(x,y,z)=k\rho'\iiint_V\frac{\hat{\mathbf{r}}}{(x'-x)^2+(y'-y)^2+(z'-z)^2}\,dx'\,dy'\,dz'\,$$ where $$\hat{\mathbf{r}}$$ is the unit vector pointing from $$(x',y',z')$$ to $$(x,y,z)$$, that is, $$\hat{\mathbf{r}}=\frac{(x-x')\hat{\mathbf{i}}+(y-y')\hat{\mathbf{j}}+(z-z')\hat{\mathbf{k}}}{\sqrt{(x'-x)^2+(y'-y)^2+(z'-z)^2}}$$ The electric field at a point charge is infinite. Therefore, I don't think you will get a convergent integral over that "Gaussian surface" because that's where you are directly at the charges sitting on the boundary of the cube.