SciLearn: electric field integral

Monday, September 8, 2008

electric field integral

During my physics class, I had an encounter with a tough integral. I was trying to derive the electric field for an infinite line of charges, but got stuck because I couldn't integrate:
$\int{\mathrm{d}x \over (x^2 + y^2)^{3 \over 2} }$ (y is a constant)
There are many sources about how the integral can be integrated, but none of them (as far as I have found) indicate how the final result was obtained. So here's one way of deriving the answer (credit: HallsofIvy at Physics Forums):
$\begin{align*} & \int{\mathrm{d}x \over (x^2 + y^2)^{3 \over 2} } \\&= \int{\mathrm{d}(y \tan \theta) \over [(y \tan \theta)^2 + y^2]^{3 \over 2}} \\&= \int{y \sec^2 \theta \,\mathrm{d}\theta \over [y^2 (\tan^2 \theta + 1)]^{3 \over 2}} \\&= \int{y \sec^2 \theta \,\mathrm{d}\theta \over y^3 \sec^3 \theta} \\&= \int{\mathrm{d}\theta \over y^2 \sec \theta} \\&= {1 \over y^2} \int \cos \theta \,\mathrm{d}\theta \\&= {1 \over y^2} \sin \theta + C \\&= {1 \over y^2} {x \over \sqrt{x^2 + y^2}} + C \\&= {x \over y^2 \sqrt{x^2 + y^2}} + C \end{align*}$
Okay, just a brief explanation:
In the 2^nd equation, substitute x = y tan θ and dx = y sec² θ
In the 4^th equation, simplify the equation with tan² θ + 1 = sec² θ
In the 8^th equation, simplify with:
$\begin{align*} \textup{since} \qquad x &= y \tan \theta \\ x &= y {\sin \theta \over \cos \theta} \\ x &= y \sin \theta \sec \theta \\ x &= y \sin \theta \sqrt{\sec^2 \theta} \\ x &= y \sin \theta \sqrt{\tan^2 \theta + 1} \quad \Leftarrow \textup{from }\sec^2 \theta = \tan^2 \theta + 1 \\ x &= \sin \theta \sqrt{y^2 \left(\tan^2 \theta + 1\right)} \\ x &= \sin \theta \sqrt{\left(y \tan \theta\right)^2 + y^2} \\ x &= \sin \theta \sqrt{x^2 + y^2} \qquad \Leftarrow \textup{since }x = y \tan \theta \\ {x \over \sqrt{x^2 + y^2}} &= \sin \theta \end{align*}$