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Surface Integral

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Surface Integral Surface integral is a definite integral taken over a surface. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate over its scalar fields and vector fields. Surface integrals have applications in physics, particularly with the classical theory of electromagnetism. The definition of surface integral relies on splitting the surface into small surface elements.

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Surface Integral In line integral, we learned how to integrate along a curve. We will now learn how to perform integration over a surface in R3, such as a sphere or a paraboloid. Similar to how we used a parametrization of a curve to define the line integral along the curve, we will use a parametrization of a surface to define a surface integral. We will use two variables, u and v, to parametrize a surface G in R3 : x = x(u,v), y = y(u,v), z = z(u,v), for (u,v) in some region R in R2

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Surface Integral

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Surface Integral In this case, the position vector of a point on the surface G is given by the vector-valued function

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Surface Integral Now take a point (u,v) in R as, say, the lower left corner of one of the rectangular grid sections in R. Suppose that this rectangle has a small width and height of Δu and Δv, respectively. The corner points of that rectangle are (u,v), (u + Δu,v), (u+Δu,v+Δv) and (u,v+Δv). So the area of that rectangle is A = ΔuΔv. Then that rectangle gets mapped by the parametrization onto some section of the surface G which, for Δu and Δv small enough, will have a surface area (call it dS) that is very close to the area of the parallelogram which has adjacent sides r(u+ Δu,v) − r(u,v) (corresponding to the line segment from (u,v) to (u + Δu,v) in R) and r(u,v+ Δv) − r(u,v)

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**Surface Integral We have**

so the surface area element dσ is approximately

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**Surface Integral Definition**

Let G be a surface in R3 parametrized by x = x(u,v), y = y(u,v), z = z(u,v), for (u,v) in some region R in R2. Let r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k be the position vector for any point on G, and let f(x, y, z) be a real-valued function defined on some subset of R3 that contains G. The surface integral of f(x, y, z) over G is In particular, the surface area S of G is

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Surface Integral In special case of a surface G described as a graph z = h(x, y) of a function h defined on a region R in the xy-plane, we may use x and y (rather than u and v) as the parameters. The surface integral of f(x, y, z) over G is

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Example 1 A torus T is a surface obtained by revolving a circle of radius a in the yz-plane around the z-axis, where the circle’s center is at a distance b from the z-axis (0 < a < b), as in figure. Find the surface area of T.

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Solution the torus can be parametrized as:

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Solution

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Example 2 Evaluate where G is the part of the plane 2x – y + z = 3 above the triangle R with vertices (0, 0), (1, 0), and (1, 1)

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**Flux of a Vector Field Through a Surface**

Let G be such a smooth, two sided surface, and assume that it is submerged in a fluid with a continuous velocity field F(x,y,z). If ΔS is the area of a small piece of G, then F is almost constant there, and the volume ΔV of fluid crossing this piece in the direction of the unit normal n is ΔV F . n ΔS We conclude that Flux of F across G =

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Example 3 Find the upward flux of F = – y i + x j + 9 k across the part of the spherical surface G determined by

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**Flux of a Vector Field Through a Surface**

Theorem Let G be a smooth, two sided surface given by z = f(x,y), where (x, y) is in R, and let n denote the upward unit normal on G. If f has continuous first order partial derivatives and F = M i + N j + P k is a continuous vector field, then the flux of F across G is given by

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**Flux of a Vector Field Through a Surface**

Proof: If we write H(x, y, z) = z – f(x, y), we obtain It follows from definition of surface integral that

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**Exercise Evaluate the surface integral , where**

F(x, y, z) = yz i + xz j + xy k and G is the part of the plane x + y + z = 1 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n pointing in the positive z direction (see figure)

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Exercise 2. Calculate the flux of F = y i – x j + 2 k across G where G is the surface determined by 3.

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