关于这个问题，设曲面方程为 $z=f(x,y)$，则曲面面积元素为 $dS=\\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}dxdy$。

则曲面面积为：$$S=\\iint_D \\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}dxdy$$利用重积分的概念，可以将上式写成：$$S=\\iint_D \\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}dxdy=\\iint_D \\frac{\\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}}{\\sqrt{1}}dxdy$$将分母 $\\sqrt{1}$ 乘上 $\\frac{\\partial z}{\\partial z}$，得：$$S=\\iint_D \\frac{\\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}}{\\sqrt{1+(\\frac{\\partial z}{\\partial x})^2+(\\frac{\\partial z}{\\partial y})^2}}(\\frac{\\partial z}{\\partial z})dxdy$$上式中的 $\\frac{\\sqrt{1+(\\frac{\\partial f}{\\partial x})^2+(\\frac{\\partial f}{\\partial y})^2}}{\\sqrt{1+(\\frac{\\partial z}{\\partial x})^2+(\\frac{\\partial z}{\\partial y})^2}}$ 可以看做是曲面在点 $(x,y,z)$ 处法向量的长度，即 $|\\vec{n}|$。因此，上式可以写成：$$S=\\iint_D |\\vec{n}|dxdy$$这就是利用重积分求曲面面积的公式。