Math & Science ⇒ another problem with lining up the equation

[rayman]

Hello! I have been trying to set up this equation in many ways but it does not look very nice...Can we move these 4 rows a bit to the left so they are more less in the center? It would be great to move them so they all lie under the second integral in the first row. Can someone smart help me?

[Stefan Kottwitz]

Put the & before the = in the first row, and before the alignment place in the other rows. Perhaps post your code for that equations here, for seeing how you did it and for fixing.

Stefan

[rayman]

Put the & before the = in the first row, and before the alignment place in the other rows. Perhaps post your code for that equations here, for seeing how you did it and for fixing.

Stefan

Oh yes, I totally forgot here is my latex code

Code: Select all

\int\limits_{S}f=\int\limits_{S}\frac{1}{2}f_{\mu\nu}(x(u))J^{\mu\nu}(x(u))du^{1}\wedge du^{2}\\
               &=\int\limits_{S}\frac{E}{4\pi}\frac{Rsin\theta cos\phi}{R^3}J^{yz}(\theta,\phi)d\theta\wedge d\phi+\\
               &+\int\limits_{S}\frac{E}{4\pi}\frac{Rsin\theta sin\phi}{R^3}J^{zx}(\theta,\phi)d\theta\wedge d\phi\\
               &+\int\limits_{S}\frac{E}{4\pi}\frac{Rcos\theta}{R^3}J^{xy}(\theta,\phi)d\theta\wedge d\phi\\
               &=\frac{E}{4\pi}\int_{0}^{2\pi}d\theta\int_{0}^{\pi}d\theta sin\theta=-\frac{E}{2}cos\theta|_{0}^{\pi}\\
               &=E

P.S Why latex does not work here?

[Stefan Kottwitz]

P.S Why latex does not work here?

Just use code instead of latex, I edited it. I'll have a look.

Stefan

[Stefan Kottwitz]

As I guessed, there's just a & missing before the equal sign in the first row. Further I recommend to use operator names such as \sin and \cos:

Code: Select all

\begin{align*}
\int\limits_{S}f&=\int\limits_{S}\frac{1}{2}f_{\mu\nu}(x(u))J^{\mu\nu}(x(u))du^{1}\wedge du^{2}\\
               &=\int\limits_{S}\frac{E}{4\pi}\frac{R\sin\theta\cos\phi}{R^3}J^{yz}(\theta,\phi)d\theta\wedge d\phi\\
               &\quad+\int\limits_{S}\frac{E}{4\pi}\frac{R\sin\theta\sin\phi}{R^3}J^{zx}(\theta,\phi)d\theta\wedge d\phi\\
               &\quad+\int\limits_{S}\frac{E}{4\pi}\frac{R\cos\theta}{R^3}J^{xy}(\theta,\phi)d\theta\wedge d\phi\\
               &=\frac{E}{4\pi}\int_{0}^{2\pi}d\theta\int_{0}^{\pi}d\theta\sin\theta=-\frac{E}{2}\cos\theta|_{0}^{\pi}\\
               &=E
\end{align*}

align.png (11.47 KiB) Viewed 3669 times

Stefan

[rayman]

looks really nice! Many thanks!