What we’re aiming at:

∑_{i=1}^{n}(z_{i})^{3} \{∫_{x=0,y=0}^{x=π/2,y=π/2} \sqrt{(x^{2} + y^{2})} f(x, y, z_{i}) dx dy\} = \frac{A}{B}.

Build it up piece-by-piece:

• First the right-hand-side:

    \frac{\mathcal{A}}{\mathcal{B}}.

giving

\frac{A}{B}.

• Next, the factors of the integral’s kernel,

    \sqrt{(x^2 + y^2)}
    f(x,y,z_i).

yielding

\sqrt{(x^{2} + y^{2})} and f(x, y, z_{i}).

• Notice that the ‘d’ in the infinitesimals is an operator, not a variable, so

\mathrm{d}x, and put the pieces together, with appropriate spacing:

    \sqrt{(x^2 + y^2)} \, f(x,y,z_i)\,\mathrm{d}x\,\mathrm{d}y.

giving

\sqrt{(x^{2} + y^{2})} f(x, y, z_{i}) dx dy.

• Finally, the integral with bounds and curly brackets,

    \left\{    \int_{x=0,y=0}^{x=\pi/2,y=\pi/2}   \right\}

giving

\{∫{x=π/2,y=π/2 \atop x=0,y=0}\}

and the sum:

     \sum_{i=1}^{n} (z_i)^3.

yielding

∑_{i=1}^{n}(z_{i})^{3}.

Putting it all together, we have:

% -- notice the indented layout which helps us ``debug'' the equation :
\[
    \sum_{i=1}^{n} (z_i)^3
        \left\{
            \int_{x=0,y=0}^{x=\pi/2,y=\pi/2}
                \sqrt{(x^2 + y^2)} \,
                f(x,y,z_i)\,\mathrm{d}x\,\mathrm{d}y
        \right\}
%%
    = \frac{\mathcal{A}}{\mathcal{B}}
\]