Evaluating definite integral with functions as limits

Question

This was a physics example regarding electric field, but I need some help in the calculation part.

Positive charge $Q$ is uniformly distributed along the $y-axis$ between $y=-a$ and $y=+a$. Find the electric field $P$ on the $x-axis$ at a distance $x$ from the origin

So the electric field is generated by a rod, which is broke into infinitesimal segments.

Rod below

| <== $dy$

|

| -------------$x$------------ $P$

|

|

Each segment is called $dy$

$x$ is the distance from $P$ to the rod

$2a$ is the length of rod

$Q$ is total charge, $Q/2a$ is the linear charge density

After a series of steps, the textbook reached the following step, for the $x$ component of the electric field: $$ E_{x} = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{x{\,}dy}{(x^2+y^2)^{3/2}} $$

the detail of evaluation was omitted in the textbook, but it's evaluated to $$ E_{x} = \frac{Q}{4\pi \epsilon_{0}} \frac{1}{x \sqrt {x^2+a^2}} $$

I spent a good deal of time trying to evaluate this integral myself using substitution but all met dead ends. I need some guidance on this.

Answer 1

What you need is essentially the followig indefinite integral: \begin{align} \int \frac{1}{(x^2+y^2)^{3/2}}\,dy &=\int \frac{\cos u}{x^2}du& (y=x\tan(u),\ dy=x\sec^2(u)du)\\ &=\frac{\sin(u)}{x^2}+C\\ &=\frac{\sin(\tan^{-1}\frac{y}{x})}{x^2}+C\\ &=\frac{y}{x^2\sqrt{x^2+y^2}}+C \end{align}

It follows that \begin{align} \int_{-a}^a\frac{x}{(x^2+y^2)^{3/2}}dy &=2x\int_{0}^a\frac{1}{(x^2+y^2)^{3/2}}dy\\ &=2x\cdot \frac{y}{x^2\sqrt{x^2+y^2}}\big|_{y=0}^{y=a}\\ &=\frac{2a}{x\sqrt{x^2+a^2}} \end{align}

Answer 2

First, notice that $x$ is a constant with respect to the variable of integration, and so

\begin{align} E_{x} &= \frac{1}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{x{\,}dy}{(x^2+y^2)^{3/2}}\\ & = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}}. \end{align}

Now, let \begin{align} y &= x\tan(u)\\ dy &= x\sec^{2}(u)\,du. \end{align}

Then, using the identity $\tan^2(u) + 1 = \sec^2(u)$, we obtain \begin{align} \int \frac{dy}{(x^2+y^2)^{3/2}} &=\int\frac{x\sec^{2}(u)\,du}{(x^{2} + x^{2}\tan^{2}(u))^{3/2}}\\ &=\int\frac{x\sec^{2}(u)\,du}{(x^{2}\sec^2(u))^{3/2}}\\ &=\frac{1}{x^{2}}\int \cos(u)\,du\\ &=\frac{1}{x^{2}}\sin(u) + C. \end{align}

Now, $$\tan(u) = \frac{y}{x} \implies \sin(u) = \frac{y}{\sqrt{x^2 + y^2}},$$ and so we have $$\int \frac{dy}{(x^2+y^2)^{3/2}} = \frac{y}{x^{2}\sqrt{x^{2} + y^{2}}} + C.$$ Using this to evaluate the definite integral: $$\int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}} = \frac{y}{x^{2}\sqrt{x^{2} + y^{2}}}\Bigg|_{-a}^{a} = \frac{2a}{x^{2}\sqrt{x^2 + a^2}}.$$ Putting everything back together, then, we obtain $$E_{x} = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}} = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a}\frac{2a}{x^{2}\sqrt{x^2 + a^2}}=\frac{Q}{4\pi \epsilon_{0}} \frac{1}{x \sqrt {x^2+a^2}}.$$