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Surface Integral

Surface Integral. Surface Integral. S urface 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 .

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Surface Integral

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  1. Surface Integral

  2. 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.

  3. Surface Integral • In line integral, we learned how to integrate along a curve. We will now learn how to performintegration 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 thecurve, 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

  4. Surface Integral

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

  6. 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 widthand 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 thearea 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)

  7. Surface Integral • We have • so the surface area element dσis approximately

  8. Surface Integral • Definition Let G be a surface in R3parametrized 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 theposition vector for any point on G, and let f(x, y, z) be a real-valued function defined on somesubset 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

  9. 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

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

  11. Solution • the torus can be parametrized as:

  12. Solution

  13. 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)

  14. 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 . nΔS • We conclude that Flux of F across G =

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

  16. 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

  17. 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

  18. 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)

  19. 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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