The formula that I have been using for finding area is of a surface is :
$A = \iint dS = \iint \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 } \space dxdy$
So I tried to set this problem as so :
$A = \iiint dS = \iiint \sqrt{1 + (\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2 } \space dxdydz$
In this question setting the bounds and finding the partial derivatives lead me to this :
$\int_{0}^{1} \int_{0}^{z} \int_{-\sqrt{z^2-y^2}}^{\sqrt{z^2-y^2}} \sqrt{1 + 4x^2 + 4y^2 + 4z^2} dxdydz$
This gives an answer in wolfram alpha that is equivalent to the answer in the text but the text gives this formula for finding the surface area involving three variables :
$A = \iint dS = \iint \sqrt{1 + (\frac{\partial (y,z)}{\partial (u,v)})^2 + (\frac{\partial (x,z)}{\partial (u,v)})^2 + (\frac{\partial (x,y)}{\partial (u,v)})^2} \space dudv$
Why is it neccesary to paramatize the surface by u and v then take the cross product of each ?