<s>[INST] How can I expand on this part of this proof concerning magnetic fields? I am supposed to prove:
$$B_y=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi}  \frac{\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3yz}{r^2}\biggl)$$
and
$$B_z=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi}  \frac{a-y\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3z^2}{r^2}-1\biggl)$$
One of the steps to doing it is proving:
$$(a^2+y^2+z^2-2ay\sin\phi)^{3/2}\approx r^{-3}\biggl(1-\frac{2ay\sin\phi}{r^2}\biggl)^{-3/2}\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl)$$
using the Binomial theorem (the specific instruction reads "for this step use the binomial theorem, expand up to second term"). How can I go about this step, taking into account the following?
$$r=\sqrt{z^2+y^2}>>a$$
 [/INST] First note that:

*

*$r = \sqrt{z^2+y^2} \gg a \implies z^2+y^2 \gg a^2 \implies a^2+y^2+z^2 \approx y^2+z^2$


*$r \gt y \;\land\; r \gg a \implies r^2 \gg ay \implies r^2 \gg ay \sin \phi$
Then:
$$
\require{cancel}
\begin{align}
(a^2+y^2+z^2-2ay\sin\phi)^{-3/2} &\approx (y^2+z^2-2ay\sin\phi)^{-3/2}
\\ &= \left(r^2 \left(1-\frac{2ay\sin\phi}{r^2}\right)\right)^{-3/2}
\\ &= \left(r^{2}\right)^{-3/2} \cdot \left(1- \left(-\frac{3}{\cancel{2}}\right) \cdot \frac{\cancel{2}ay\sin\phi}{r^2} + \dots\right)
\\ &\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl)
\end{align}
$$
</s>