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Chapter 6 Multiple Integral

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Chapter 6 Multiple Integral Chapter 6 Multiple Integral 157 6.1 Introduction Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. (Note that the same volume can be obtained via the triple integral. If there are more variables, a multiple integral will yield hypervolumes of multi-dimensional functions. Multiple integration of a function in n variables: f(x1, x2, ..., xn) over a domain D is most commonly represented by nested integral signs in the reverse order of execution (the leftmost integral sign is computed last), followed by the function and integrand arguments in proper order (the integral with respect to the rightmost argument is computed last). The domain of integration is either represented symbolically for every argument over each integral sign, or is abbreviated by a 158 variable at the rightmost integral sign. In this chapter, we will discuss line, double, triple integrals. 6.2 Line integral In mathematics, a line integral (sometimes called a path integral, contour integral, or curve integral; not to be confused with calculating arc length using integration) is an integral where the function to be integrated is evaluated along a curve. The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighing distinguishes the line integral from simpler integrals defined on intervals. The line integral of f(x,y) along C is denoted by f(x,y) ds , c We use a ds here to acknowledge the fact that we are moving along the curve, C, instead of the x-axis (denoted by dx) or the y-axis (denoted by dy). Because of the ds this is 159 sometimes called the line integral of f with respect to arc dy length such that ds = (dx )2 +( ) 2 dt. dt dt Parametric form of line segment joining 2 points (x0, y0, z0) to (x1, y1, z1) is expressed by: x = (1-t) x0 + t x1, y = (1-t) y0 + t y1 , z = (1-t) z0 + t z1, 0<t< 1 Line integral of vector field In this section we are going to evaluate line integrals of vector fields. We’ll start with the vector field, F(x,y,z)= P(x,y,z)i Q(x,y,z)j R(x,y,z)k And the three-dimensional, smooth curve given by r(t)= x(t)i y(t)j z(t)k , a ≤ t ≤ b Then the line integral of F along c is b F dr F( r (t)) r (t) dt c a 160 That really is a dot product of the vector field and the differential and the differential really is a vector. Also F(r(t)) F(x(t),y(t),z(t)) We can also write line integrals of vector fields as a line integral with respect to arc length as follows, b F dr F T ds c a r (t) Where T(t) is the unit vector and is given by T(t)= r (t) If we use our knowledge on how to compute line integrals with respect to arc length we can see that this second form is equivalent to the first form given above b r (t) b F dr Fr(t))(( r(t)dt Fr(t)) r(t)dt c ar (t) a 161 Example 1 Evaluate F dr , where F(x,y,z) 8x2 yz i 5z j 4xyk and c c is the curve given by r(t)= t i t2 j tk 3 , 0 t 1. Solution: We first need the vector field evaluated along the curve. F( r (t)) 8t7 i 5t 3 j 4t 3 k , r (t) i 2t j 3t2 k , then F( r(t)) r (t) = 8t7 10t 4 12t 5 , therefore: 1 b F dr F( r(t)) r (t)dt = [8t7 10t 4 12t 5 ]dt = 1 c a 0 Example 2 2 2 Evaluate xy4 ds , c is the right half of the circle x + y = 16 c rotated in the counter clockwise direction. 162 Solution: Since the parametric equation of the circle is x = 4cost, y = 4sint, then dx = -4sint and dy = 4cost, t , dt dt dy therefore (dx )2 +( ) 2 = 16 and ds = 4 dt. dt dt The line integral is then xy4 ds (4cost)(4sint) 4 (4)dt c - = 8192/5. The line integral can be denoted by P(x,y) dx + Q(x, y) dy . c To integrate this form, we have to express it in one variable (x or y or parametric equation) using the contour equation. Example 3 Evaluate sin( y)dx+yx2 dy , c is line segment from (0, 2) to c (1, 4). 163 Solution: Since the contour integration is the line joining (0,2) and (1,4) such that y = 2x + 2, therefore dy = 2dx. 1 Hence sin( y)dx+yx2 dy sin(2 x)dx+(2x+2)x 2 (2)dx c 0 1 cos(2 x) 4 1 7 sin(2 x)dx+4(x3 +x 2 )dx (x4 x 3 ) 0 2 3 0 3 Example 4 Evaluate sin( y)dx+yx2 dy , c is line segment from (1,4) to c (0,2) Solution: Since the contour integration is the line joining (1,4) and (0,2) such that y = 2x + 2, therefore dy = 2dx. 0 Hence sin( y)dx+yx2 dy sin(2 x)dx+(2x+2)x 2 (2)dx c 1 164 0 0 3 2 cos(2 x)4 43 7 [sin(2 x)+4(x +x )]dx (x x ) 1 2 31 3 From above example we note that P(x,y)dx+Q(x,y)dy P(x,y)dx+Q(x,y)dy c -c Example 5 Evaluate ydx+xdy+zdz, c is given by x = cost, y = sint, c z = t2, where 0 < t < 2. Solution: Since dx = -sint dt, dy = cost dt, dz = 2t dt, therefore: 2 ydx+xdy+zdz (sint(-sint)+cost(cost) +t2 (2t)) dt c 0 2 sin 2t t4 2 = (cos2t+2t)dt3 ( ) 8 4 0 2 2 0 In any region where P dx + Q dy is independent of the path, the partial derivatives of the function (x,y) x,y)= P(x,y)dx+Q(x,y)dy (a,b) 165 are P(x,y) & Q(x,y) . xx y y Theorem If QP at all points of a simply connected region R, then x y the integral P(x,y) dx + Q(x,y) dy is independent on the path and conversely. Example 6 (2,3) Evaluate the integral (3x2 3y)dx+(3x-4y)dy , along (-1,0) the path y2 – 2 x2y + x4 -1 = 0. Solution: Since QP3 , Q(x, y) = 3x - 4y, P(x, y) = 3x2 3y , x y therefore the integration independent on the path, so we can consider the line y = x+1 is our path such that dy = dx and (2,3) (3x2 3y)dx+(3x-4y)dy (-1,0) 166 2 = (3x2 3(x+1))dx +(3x-4(x+1))dx -1 2 = (3x2 2x-1)dx=9 -1 Problems Evaluate the following integrals 1-(xy + ln x) dy, c is the arc of parabola y = x2 from (1,1) to c (3,9) 2- (x+2y)dx+(x-y)dy , c is the curve given by x = 2cost, c y = 4sint 3-4x3 ds, c is the line segment from (-2,-1) to (1,2) c 4- x ds, c is the curve given by y = x2, -1< x < 1 c 5- xyz ds, c is the helix given by r = (cost, sint, 3t), 0≤ t≤ 4 c 6- F dr , where F(x,y,z)= xz i yz k and c is the line c segment from (-1,2,0) and (3,0,1) ########################################################## 167 6.3 Green theorem In this section we are going to investigate the relationship between certain kinds of line integrals (on closed paths) and double integrals. Let’s start off with a simple (recall that this means that it doesn’t cross itself) closed curve C and let D be the region enclosed by the curve. Here is a sketch of such a curve and region. C D Let C be a positively oriented, piecewise smooth, simple, closed curve and let D be the region enclosed by the curve. If P and Q have continuous first order partial derivatives on D QP then: P(x,y) dx + Q(x,y) dy ( ) dxdy c D x y 168 Example 7 Use Green’s Theorem to evaluate xy dx + x2 y 3 dy, where c c is the triangle whose vertices (0,0), (1,0), (1,2) with positive orientation Solution y (1,2) I3 I2 (0,0) I1 (1,0) x If we express the problem in line integral, we have to divide the path into three paths I1 : y = 0, I2 : x = 1, I3 : 2x = y so that dy = 0, dx = 0, 2dx = dy respectively. 1 2 3 2 3 For path I1: xydx+x ydy xydx+x ydy=0 I 0 1 2 2 3 3 For path I2: xydx+xydy ydy=4 , I2 0 0 2 3 2 3 10 For path I3: xydx+x y dy [x(2x)+x (2x) (2)]dx .
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