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  "body": "\n## Definition\n\nLet $r \\colon D \\to \\mathbb{R}^3$ be a [[parametrised surface|05788J51WZ]] as in the definition of [[surface area|0WQE3G7B88]], let $S = r(D)$, and let $f \\colon S \\to \\mathbb{R}$ be continuous.\nThe surface integral of $f$ over $S$ is\n$$\\iint_S f \\, dS = \\iint_D f\\big( r(u, v) \\big) \\, \\| r_u(u, v) \\times r_v(u, v) \\| \\, du \\, dv.$$\n\n## Remarks\n\nThe expression $dS = \\| r_u \\times r_v \\| \\, du \\, dv$ is the surface element. With $f = 1$, the integral is the area of $S$.\nThe value does not depend on the parametrisation of $S$. For a surface made of several pieces, add the integrals over the pieces.\nFor the graph $z = g(x, y)$ over $D$, $\\iint_S f \\, dS = \\iint_D f(x, y, g(x, y)) \\sqrt{1 + g_x^2 + g_y^2} \\, dx \\, dy$.\nExample: on the unit sphere $S$, symmetry gives $\\iint_S x^2 \\, dS = \\iint_S y^2 \\, dS = \\iint_S z^2 \\, dS$. These add up to the area $4\\pi$, so $\\iint_S z^2 \\, dS = 4\\pi / 3$.\n\n## Explanations\n\n### L1 human\n\nA surface integral adds up the values of a function over a curved surface, each weighted by a small piece of area.\nIf $f$ is a mass density per unit area, the surface integral is the total mass of the surface.\n",
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