51. General Surface Integrals

Transcription

1 51. General urface Integrals The area of a surface in defined parametrically by r(u, v) = x(u, v), y(u, v), z(u, v) over a region of integration in the input-variable plane is given by d = r u r v da. Now, let w = f(x, y, z) be a function defined over this surface. We then wish to calculate the surface integral, where f(r(u, v)) = f(x(u, v), y(u, v), z(u, v)): f(r(u, v)) d = f(r(u, v)) r u r v da. When the surface is defined explicitly by a function z = g(x, y), then r(x, y) = x, y, g(x, y), and the surface integral can be rewritten f(x, y, z) d = f(x, y, g(x, y)) (g x (x, y)) + (g y (x, y)) + 1 da, Where d = r u r v da = (g x (x, y)) + (g y (x, y)) + 1 da. urface area integrals are a special case of surface integrals, where f(x, y, z) = 1. urface integrals can be interpreted in many ways. ome examples are discussed at the end of this section. Example 51.1: Find (x + yz) d, where is the surface z = 1 4x y contained in the first quadrant. olution: Here, z = g(x, y) = 1 4x y, so that g x = 4 and g y =. Thus, d is d = ( 4) + ( ) + 1 = 6 da. The integrand is written in terms of x and y, with the substitution z = 1 4x y: x + yz = x + y(1 4x y) = x + 1y 4xy y

2 The region of integration is the footprint of the surface projected onto the xy-plane. Below is a sketch of and its region of integration. Letting da = dy dx, we have y 4 x + 4 and x as the bounds of : The surface integral is now (x + yz) d = (x + 1y 4xy y ) 6 (4 )x+4 da = 6 (x + 1y 4xy y ) dy dx. The inside integral is (4 )x+4 (x + 1y 4xy y ) dy = [xy + (6 x)y y (4 )x+4 ] Note that the two middle terms, 1y 4xy, can be written (1 4x)y, which gives (6 x)y after integration with respect to y. ubstituting and simplifying, we obtain x ( 4 x + 4) + (6 x) ( 4 x + 4) ( 4 x + 4) This is now integrated with respect to x: 6 ( 7 x + 8 x 8x + ) dx = 7 x + 8 x 8x +. = 6 [ 8 7 x x 14x + x] = 6.

3 Example 51.: Find x d, where is the portion of sphere of radius 4, centered at the origin, such that x and z. olution: The surface is a quarter-sphere bounded by the xy and yz planes. We sketch and from it, infer the region of integration : The hemisphere can be described by rectangular coordinates x + y + z = 16, in which case z = g(x, y) = 16 x y. From this, we obtain partial derivatives Thus, z x = x d = ( 16 x y ) x y and z y = 16 x y 16 x y. y + ( 16 x y ) + 1 da = 4 16 x y da. (ee Example 5. for a similar example with all steps shown.) Using polar coordinates, where x = r cos θ, y = r sin θ, and r = x + y, the region is described by r 4 and π θ π. Due to the change of variables, the differential element is now da = r dr dθ. The surface integrals now is written x d 4 = (r cos θ) 4 r dr dθ. 16 r 4

4 This simplifies to To integrate Thus, r 16 r 4 4 ( r cos θ dr dθ. 16 r), we use the form taken from a table of integrals: r a r dr = 1 (a + r ) a r. 4 r 16 r dr = [ 1 4 ( + r ) 16 r ] = 18. We now evaluate the outer integral, using the identity cos θ = cos(θ). Note that the constant 18 moves to the front of the integral: 4 ( 18 ) cos θ dθ = 51 (1 + 1 cos θ) dθ = 56 (1 + cos θ) dθ = 56 [θ + 1 sin θ] = 56 [(π + 1 sin(π)) ( π + 1 sin( π))] = 56 π. In the next example, we revisit the previous example using spherical coordinates. 5

5 Example 51.: Find x d, where is the portion of sphere of radius 4, centered at the origin, such that x and z. Parameterize using spherical coordinates. olution: Using spherical coordinates and the usual parameterization of a sphere with a fixed radius, we have r(φ, θ) = 4 sin φ cos θ, 4 sin φ sin θ, 4 cos φ, where π θ π and φ π. From this, we determine r φ r θ. The derivation is lengthy but not difficult, as many trigonometric identities can be used to simplify. ee Example 5.4 for one such example. We have Thus, the surface integral is x d r φ r θ = 16 sin φ. = (4 sin φ cos θ) 16 sin φ dφ dθ. where x = (4 sin φ cos θ) and d = r φ r θ da = 16 sin φ dφ dθ. This simplifies to 56 sin φ cos θ dφ dθ. Because the bounds are constant and the integrand held by multiplication, we can rewrite the integral as 56 sin φ cos θ dφ dθ = 56 ( sin φ dφ ) ( cos θ dθ Both require some techniques of trigonometric integration. For the integrand cos θ, we use the identity cos θ = cos(θ): cos θ dθ = ( cos(θ)) dθ = [ 1 θ sin(θ)] = ( π 4 + ) ( π 4 + ) = π. ). 6

6 For the integrand, sin φ is rewritten as sin φ sin φ = (1 cos φ) sin φ: sin φ dφ = sin φ sin φ dφ = (1 cos φ) sin φ dφ = sin φ cos φ sin φ dφ = [ cos φ + 1 cos φ] = ( ) =. Assembling this information together, we have x d = 56 ( sin φ dφ = 56 ( ) (π ) = 56 π. ) ( cos θ dθ You can decide if this method is more efficient than using rectangular coordinates. Clearly, both methods work. It is important to observe that we did not include the Jacobian ρ sin φ when we developed the integral in the previous example (Example 51.) in variables φ and θ. This is because we originally parameterized the surface in φ and θ, in which case, the area differential elements will always be da = du dv, or da = dφ dθ in this case. emember, the derivation of d = r φ r θ da does not know whether the variables represent rectangular, spherical or cylindrical coordinate systems. However, if we parameterize the surface in generic variables u and v, and then midway through the problem decide to integrate with respect to a different coordinate system, then we must include the Jacobian when we convert the area differential element. We saw this in the example prior to the last one (Example 51.). ) 7

7 etting up a surface integral is usually not difficult. However, in many cases, the integrands can be difficult to antidifferentiate. A computer, tables of integrals or numerical methods may need to be used. This is shown in the next example. Example 51.4: Find xy-plane. x z d, where is the paraboloid z = g(x, y) = 1 x y over the olution: From the surface, we have partial derivatives g x = x and g y = y. Thus, The surface integral is d = ( x) + ( y) + 1 da = 4x + 4y + 1 da. x z d = x (1 x y ) 4x + 4y + 1 da. This is a difficult integral to evaluate if we remain in rectangular coordinates. Thus, we convert to polar coordinates, where the region of integration is a circle of radius 1, centered at the origin on the xy-plane: x (1 x y ) 4x + 4y + 1 da = Using a computer or numerical methods, Meanwhile, Thus, π π 1 (r cos θ) (1 r ) 4r + 1 r dr dθ π 1 (r r 5 ) 4r + 1 dr 1 = cos θ (r r 5 ) 4r + 1 dr dθ,.14, cos θ dθ = π (using a trigonometric identity such as in the previous example). x z d.14π. 8

8 Applications of urface Integrals There are a handful of common applications of surface integrals that may help one intuitively understand them better. For example, if z = g(x, y) is a surface with uniform thickness, and f(x, y, z) represents the density at each point (x, y, z) on the surface, then be interpreted as the total mass of. f(x, y, z) d can From Example 51.1, we had (x + yz) d, where was the surface z = 1 4x y contained in the first quadrant. If x, y and z are measured in meters, and f(x, y, z) = x + yz is the density of the object at the point (x, y, z) in kilograms per square meter, then (x + yz) d is the total mass of the surface, in kilograms. We could interpret the result by claiming that this surface has a total mass of 6 kilograms. In Example 51., suppose that the integrand f(x, y, z) = x represents the density of a population of bacteria (in thousands per square centimeter) on the surface at any given point (x, y, z), where the variables are measured in centimeters. Then, the total population would be given by x d = 56 π 68, or about 68, bacteria. Furthermore, the average density of the object represented by the surface would be its total mass divided by the surface s area: Average density = f(x, y, z) d d. In Example 51.1, the surface area of is of the object is d = 6 6 square meters. Thus, the average density = 5 kilograms per square meter. In Example 51., the surface area of the quarter-sphere is d = 1 4 (4 π(4) ) = 64 π, so the average density of the bacteria on this surface is (56 )π = 4, or 4, bacteria per square centimeter. (64 )π urface integrals are also used to find the flow of material through a surface, discussed in ection 5, Flux Integrals. 9