surface area of implicit surface

Question

Ok, so I read ( on Thomas' book) that the surface area of an implicit surface is the double integral of the magnitude of the function's gradient divided by the magnitude of the dot product between such gradient and the unit normal vector p over the projection of this surface onto a plane. The p vector would be represented by the i, j or k vector, depending on which plane we projected the surface. The problem is... I don't understand this at all. Like, I get that for an implicit surface the gradient is normal to it, but I don´t get why we would integrate the gradient, nor why there is a dot product in the middle of all of this, between the gradient and p. Could someone explain it to me, or give me some intuition? Thanks

Answer 1

If you project a piece $S$ of a planar region in the plane with unit normal vector $\vec n$ onto a region $R$ in the $xy$-plane (with normal vector $\vec k$), then it's a little bit of geometry to see that $$\text{area}(S) = \frac1{|\cos\gamma|}\text{area}(R) = \frac1{|\vec n\cdot\vec k|}\text{area}(R).$$ (Here $\gamma$ is the angle between the two planes, which coincides with the angle between the two unit normal vectors.) At a point of your implicit surface $f(x,y,z)=\text{constant}$, the unit normal vector is $\vec{\nabla f}/\|\vec{\nabla f}\|$, and so the same formula works for a little bit of surface area, approximating the little bit of the surface by a little region in the tangent plane. Note that $$\frac1{|\vec n\cdot \vec k|} = \frac1{\left|\frac{\vec{\nabla f}}{\|\vec{\nabla f}\|}\cdot\vec k\right|} = \frac{\|\vec{\nabla f}\|}{|\vec{\nabla f}\cdot\vec k|}.$$