# Surface Integral.

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Surface Integral

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

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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.

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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

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Solution 1: = surface B• dS = surface Bx dy dz
= 3 z=0z=2 y=0 y=3 yzdydz = 3y2/2 |30 z=2z=0zdz = 3×9/2 z=2z=0zdz = 3×9/2 z2/2 |20 = 27 liters min-1

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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

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

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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

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Conversion of Coordinates
Rectangular Circular Spherical x y z r z r cos -sin 0 sincos coscos -sin x y z 1 0 0 0 1 0 sin cos 0 sinsin cossin cos 0 0 1 0 0 1 Cos sin 1 0 0 r z cos sin -sin cos 0 1 0 -sin cos 0 0 1 cos -sin 0 0 1 r sincos sinsin cos sin cos 1 0 0 coscos cossin -sin cos sin 0 1 0 -sin cos 0 0 1

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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

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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

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

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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

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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

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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:

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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:

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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.

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

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