09/06/05 Example The Surface Integral.doc 2/5 Jim Stiles The Univ. The terms path integral, curve integral, and curvilinear integral are also used. C. Surface Integrals Double Integrals A function Fx y ( , ) of two variables can be integrated over a surface S, and the result is a double integral: ∫∫F x y dA (, ) (, )= F x y dxdy S ∫∫ S where dA = dxdy is a (Cartesian) differential area element on S.In particular, when Fx y (,) = 1, we obtain the area of the surface S: A =∫∫ S dA = ∫∫ dxdy We will define the top of the cylinder as surface S 1, the side as S 2, and the bottom as S 3. Some examples are discussed at the end of this section. of Kansas Dept. 2 Surface Integrals Let G be defined as some surface, z = f(x,y). Surface Integrals in Scalar Fields We begin by considering the case when our function spits out numbers, and we’ll take care of the vector-valuedcaseafterwards. Use the formula for a surface integral over a graph z= g(x;y) : ZZ S FdS = ZZ D F @g @x i @g @y j+ k dxdy: In our case we get Z 2 0 Z 2 0 Parametric Surfaces – In this section we will take a look at the basics of representing a surface with parametric equations. In this situation, we will need to compute a surface integral. The surface integral will therefore be evaluated as: () ( ) ( ) 12 3 ss1s2s3 SS S S and integrate functions and vector fields where the points come from a surface in three-dimensional space. Surface area integrals are a special case of surface integrals, where ( , , )=1. 8.1 Line integral with respect to arc length Suppose that on … The surface integral is defined as, where dS is a "little bit of surface area." In order to evaluate a surface integral we will substitute the equation of the surface in for z in the integrand and then add on the often messy square root. Often, such integrals can be carried out with respect to an element containing the unit normal. Solution In this integral, dS becomes kdxdy i.e. the unit normal times the surface element. These integrals are called surface integrals. 1 Lecture 35 : Surface Area; Surface Integrals In the previous lecture we deflned the surface area a(S) of the parametric surface S, deflned by r(u;v) on T, by the double integral a(S) = RR T k ru £rv k dudv: (1) We will now drive a formula for the area of a surface deflned by the graph of a function. of EECS This is a complex, closed surface. Example 1 Evaluate the surface integral of the vector eld F = 3x2i 2yxj+ 8k over the surface Sthat is the graph of z= 2x yover the rectangle [0;2] [0;2]: Solution. 8 Line and surface integrals Line integral is an integral where the function to be integrated is evalu-ated along a curve. Example )51.1: Find ∬( + 𝑑 Ì, where S is the surface =12−4 −3 contained in the first quadrant. Example 20 Evaluate the integral Z A 1 1+x2 dS over the area A where A is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0. Soletf : R3!R beascalarfield,andletM besomesurfacesittinginR3. If f has continuous first-order partial derivatives and g(x,y,z) = g(x,y,f(x,y)) is continuous on R, then Created by Christopher Grattoni. For a parameterized surface, this is pretty straightforward: 22 1 1 C t t s s z, a r A t x x³³ ³³? After that the integral is a standard double integral Surface integrals can be interpreted in many ways. To evaluate we need this Theorem: Let G be a surface given by z = f(x,y) where (x,y) is in R, a bounded, closed region in the xy-plane. 5.3 Surface integrals Consider a crop growing on a hillside S, Suppose that the crop yeild per unit surface area varies across the surface of the hillside and that it has the value f(x,y,z) at the point (x,y,z). The surface integral will have a dS while the standard double integral will have a dA. The Divergence Theorem is great for a closed surface, but it is not useful at all when your surface does not fully enclose a solid region. 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