Surface Integrals Surface Integrals for Parametric Surfaces In the last section, we learned how to find the surface area for parametric surfaces. We cut the region in the uvplane into tiny rectangles and added up the area of the corresponding tiny parallelograms in the xyplane. The area of these parallelograms was
If we think of the surface as having varying density f(x,y,z), then the mass of this parallelogram will be
and adding up all these masses and taking the limit as the rectangle sizes approach zero, gives the definition of the surface integral.
As with finding the surface area the integral typically results in an impossible integral.
Example Find
where S is the surface r(u,v) = ui + u^{2}j + (u+ v)k 0 < u < 2 1 < v < 4 and f(x,y,z) = x + 2z
Solution We find r_{u} = i + (2u)j + k r_{v} = k and take the cross product
We have f(x(u,v),y(u,v),z(u,v)) = x(u,v) +2z(u,v) = u +2(u + v) = 3u + v We find
Although this integral is possible, its solution is quite involved. You can verify that the surface integral evaluates to approximately 525.27. Surface Integrals for Surfaces that are Functions of Two Variables We have seen before that if z = g(x,y) is a surface such that g has continuous first order partial derivatives, then the parameterization r(u,v) = ui + vj + g(u,v)k has the property that
This leads to the formula for surface integrals.
Example Find
where S is the part of the paraboloid z = x^{2} + y^{2} that lies inside the cylinder x^{2} + y^{2} = 1 and f(x,y,z) = z
Solution We have
and f(x,y,z) = z = x^{2} + y^{2} At this point, you should be thinking, "This looks like a job for polar coordinates." And we get
Let u = 1 + 4r^{2 } du = 8r dr r^{2} = 1/4 u  1/4 and the substitution gives us
Oriented Surfaces and Flux We have seen how a region R with boundary curve C can be oriented. Traveling along C, we look to see if the region is on the right or left. Unfortunately, this definition does not work will for surfaces in three dimensions. The idea of right and left is not well defined. In fact not all surfaces can be oriented. We say that a surface is orientable if a unit normal vector can be defined on the surface such that it varies continuously over the surface. Below is an example of a nonorientable surface (called the Mobius Strip)
You can see that there is no front or back of this surface. Recall that a unit normal vector to a surface can be given by
There is another choice for the normal vector to the surface, namely the vector in the opposite direction, N. By this point, you may have noticed the similarity between the formulas for the unit normal vector and the surface integral. This idea leads us to the definition of the Flux Integral Consider a fluid flowing through a surface S. The Flux of the fluid across S measures the amount of fluid passing through the surface per unit time. If the fluid flow is represented by the vector field F, then for a small piece with area DS of the surface the flux will equal to DFlux = F ^{.} N DS Adding up all these together and taking a limit, we get
Notice that the denominator of N and the formula for dS both involve r_{u} x r_{v}. Canceling, we get NdS = r_{u} x r_{v} dvdu for a surface that is defined by the function z = g(x,y), we get the nice formula NdS = g_{x}(x,y)i  g_{y}(x,y)j + k (oriented upward) or NdS = g_{x}(x,y)i + g_{y}(x,y)j  k (oriented downward)
Example Find the flux of F(x,y,z) = xi + 2yj + zk across the part of the surface z = x + y^{2} with upward pointing normal that lies within the box 0 < x < 3 2 < y < 5
Solution We compute NdS = i  2yj + k dydx and F ^{.} N dS = x  4y^{2} + x + y^{2} = 3y^{2} The flux integral is
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