Surface Integral

Surface Integral The definition of surface integral relies on splitting the surface into small surface elements.

Surface Integral It is now time to think about integrating functions over some surface, S, in three-dimensional space.  Let’s start off with a sketch of the surface S since the notation can get a little confusing once we get into it.  Here is a sketch of some surface S.

Surface Integral

Surface Integral

Surface Integral

Surface Integral

Surface Integral Problem 1: Water Flow z 0,0,2 0,3,2 y 0,0,0 0,3,0 x

Solution 1: =  surface B• dS =  surface Bx dy dz = 3 z=0z=2 y=0 y=3 yzdydz = 3y2/2 |30  z=2z=0zdz = 3x9/2  z=2z=0zdz = 3x9/2 z2/2 |20 = 27 liters min-1

Conversion of Coordinates Rectangular to Spherical Cylindrical r • X = sin cos r • X = cos r • y = sin sin r • y = sin r • z = cos r • z = 0

Conversion of Coordinates Cylindrical to Rectangular Ax = Ar cos - A sin Ay = Ar sin - A cos Az = Az

Conversion of Coordinates Spherical to Rectangular Ax = Ar sin cos + A cos cos + A sin Ay = Ar sin sin + A cos sin + A cos Az = Ar cos - A sin

Conversion of Coordinates Rectangular Circular Spherical x y z r  z r   cos -sin 0 sincos coscos -sin x y z 1 0 0 0 1 0 sin cos 0 sinsin cossin cos 0 0 1 0 0 1 Cos -sin 0 1 0 0 r  z cos sin 0 -sin cos 0 0 1 0 -sin cos 0 0 0 1 0 0 1 cos -sin 0 0 0 1 r   sincos sinsin cos sin 0 cos 1 0 0 coscos cossin -sin cos 0 -sin 0 1 0 -sin cos 0 0 1 0 0 0 1

Cross-Product of Vector The Vector product of two vectors is defined as a third vector magnitudes multiplied by the sine of the angle between them. The direction of resultant vector is perpendicular to the plane containing the first two. |C| = |A||B| sin =AB sin A x B = C = ĥ AB sin Where ĥ is unit vector normal to the plane

Cross-Product of Vector X-product of A & B z ĥ |C| = AB sin A  y B While magnitude of C is normal to both A & B x

Cross Product x x y = z y x z = x z x x = y x x x = y x y = z x z = 0 ^ ^ ^ ^ ^ ^ ^ ^ ^

Cross Product Example 1: A = x8 + y3 – z10; B = - x 15 + y6 + z17 Find A x B Example 2: A = y20 –z5 ; B = -x6 + y14

Tutorial Example 1 Evaluate where S is the portion of the zy- plane Solution : Okay, since we are looking for the portion of the plane that lies in front of the yz-plane we are going to need to write the equation of the surface in the form:           This is easy enough to do. Next we need to determine just what D is.  Here is a sketch of the surface S. Here is a sketch of the region D

Tutorial …… Notice that the axes are labeled differently than we are used to seeing in the sketch of D.  This was to keep the sketch consistent with the sketch of the surface.  We arrived at the equation of the hypotenuse by setting x equal to zero in the equation of the plane and solving for z.  Here are the ranges for y and z. Now, because the surface is not in the form   we can’t use the formula above.  However, as noted above we can modify this formula to get one that will work for us.  Here it is:

Tutorial …… The changes made to the formula should be the somewhat obvious changes.  So, let’s do the integral Notice that we plugged in the equation of the plane for the x in the integrand.  At this point we’ve got a fairly simple double integral to do.  Here is that work:

Tutorial Example 2: Evaluate Solution where S is the upper half of a sphere of radius 2 Solution We gave the parameterization of a sphere in the previous section.  Here is the parameterization for this sphere Since we are working on the upper half of the sphere here are the limits on the parameters Next, we need to determine .  Here are the two individual vectors Now let’s take the cross product.

Tutorial Example 2……. Finally, we need the magnitude of this :