surface area


A recap of the derivation of the surface area element dS. Set through the following animation to see the thought process behind finding the surface area of a surface. Each step is explained in a pop-up window: Click here to open the pop-up window for the current frame of the demonstration.

image window
Step through animation: backward one frame forward one frame

So, the key here is to find the vectors pointing along two sides of the near-parallelogram dS and take their cross product. Finding the vectors is shown in the steps given above, which arrive at the vectors

a = < dx, 0, fx dx >
and, similarly for y,
b = < 0, dy, fy dy >

(The third component of these can best be seen by thinking of the equation of a line in point-slope form: y-y0 = m (x-x0). So the change in y (in our case, z) is just the slope times the change in x -- which is dx, as shown above.)

Then the area dS = |a x b|. The cross product is

a x b = < -fx dx dy, -fy dx dy, dx dy >, so
dS = |a x b| = sqrt( (fx dx dy)2 + (fy dx dy)2 + (dx dy)2 )
= sqrt( (fx)2 + (fy)2 + 1 ) dx dy.

And, Voila!, we have the expression we want for dS!

Note! Go back and be sure that you understand this. In particular, make sure that the whole area-of-a-parallelogram thing makes sense, make sure that you see where the vectors a and b came from, do the cross-product, and make sure that you can find the magnitude to figure out what dS is!


surface areas
Last modified: Fri Mar 19 12:54:25 EST 2004
©2004 Gavin LaRose, UM Math Dept.