Line and surface integrals: Solutions

Transcription

1 hapter 5 Line and surface integrals: olutions Example 5.1 Find the work done by the force F(x, y) x 2 i xyj in moving a particle along the curve which runs from (1, ) to (, 1) along the unit circle and then from (, 1) to (, ) along the y-axis (see Figure 5.1). Figure 5.1: hows the force field F and the curve. The work done is negative because the field impedes the movement along the curve. olution plit the curve into two sections, the curve 1 and the line that runs along the y-axis 2. Then, W F dr F dr + 1 F dr. 2 urve 1 : Parameterise 1 by r(t) (x(t), y(t) (cost, sin t), where t π/2 and F (x 2, xy) and dr (dx, dy). Hence, F dr x 2 dx xydy 1 1 π/2 cos 2 t dx dt dt π/2 1 costsin t dy dt dt π/2 2 cos 2 t sintdt 2/3,

2 by applying Beta functions to solve the integral where m 2, n 1 and K 1. urve 2 : Parameterise 2 by r(t) (x(t), y(t) (, t), where t 1. Hence, 2 F dr π/2 dx dt dt π/2 t dy dt. dt o the work done, W 2/3 + 2/3. Example 5.2 Evaluate the line integral (y2 )dx+(x)dy, where is the is the arc of the parabola x 4 y 2 from ( 5, 3) to (, 2) olution Parameterise by r(t) (x(t), y(t) (4 t 2, t), where 3 t 2, since 3 y 2. F (y 2, x) and dr (dx, dy). Hence, F dr y 2 dx + xdy 2 3 t 2 dx dt dt 2 3 (4 t 2 ) dy 2 dt dt 2t 3 + (4 t 2 )dt 245/6. 3 Example 5.3 Evaluate the line integral, (x2 + y 2 )dx + (4x + y 2 )dy, where is the straight line segment from (6, 3) to (6, ). olution : We can do this question without parameterising since does not change in the x-direction. o dx and x 6 with y 3 on the curve. Hence I (x 2 + y 2 ) + (4x + y 2 )dy y 2 dy 81. Example 5.4 Use Green s Theorem to evaluate (3y esin x )dx+(7x+ y 4 + 1)dy, where is the circle x 2 + y 2 9. olution P(x, y) 3y e sin x and Q(x, y) 7x+ y Hence, Q P 7 and y 3. Applying Green s Theorem where is given by the interior of, i.e. is the disc such that x 2 + y 2 9. (3y e sin x )dx + (7x + y 4 + 1)dy (7 3)dxdy 2π 3 4rdr 2π 18 36π The integral is solved by using polar coordinates to describe. Example 5.5 Evaluate x2 (3x 5y)dx + (x 6y)dy, where is the ellipse 4 + y2 1 in the anticlockwise direction. Evaluate the integral by (i) Green s Theorem, (ii) directly. 2

3 olution (i) Green s Theorem: P(x, y) 3x 5y and Q(x, y) x 6y. Hence, Q P 1 and y 5. Applying Green s Theorem where is given by the interior of, i.e. is the ellipse such that x 2 /4+y 2 1. (3x 5y)dx + (x + 6y)dy (1 ( 5))dxdy 6 1dxdy 6 (Area of the ellipse) 6 2π. ee chapter 2 for calculating the area of an ellipse by change of variables for a double integral. (i) irectly: Parameterise by x(t) 2 cost, y(t) sin t, where t 2π. I 2π dx dy (6 cost 5 sint) dt dt + (2 cost 6 sint) dt dt 2π 18 cost sint + 1 sin 2 t + 2 cos 2 tdt + 4 π/2 sin 2 tdt + 8 π/2 cos 2 tdt + 4 π 2 (1/2) + 8 π 2 (1/2) 12π. The integrals are calculated using symmetry properties of cos t and sin t and beta functions. Using the table of signs below we see that 2π sin 2 t 4 π/2 sin t dt etc. Quadrant Total cos t + + sin t + + costsin t + + sin 2 t cos 2 t Example 5.6 Evaluate z 2 d where is the hemisphere given by x 2 + y 2 + z 2 1 with z. olution We first find etc. These terms arise because d 1 + ( )2 + ( y )2 dxdy. ince this change of variables relates to the surface we find these derivatives by differentiating both sides of the surface x 2 + y 2 + z 2 1 with respect to x, giving 2x + 2z. Hence, x/z. imilarly, y y/z. Hence, 1 + ( )2 + ( y )2 1 + x2 z 2 + y2 z 2 1/z. 3

4 Then the integrals becomes the following, where is the projection of the surface,, onto the x y-plane. i.e. {(x, y) : x 2 + y 2 1}. z 2 d z 2 1 z dxdy 1 x2 y 2 dxdy 2π 2π 2π 2π/ r2 rdr 1 1 udu 2 Example 5.7 Find the area of the ellipse cut on the plane 2x + 3y + 6z 6 by the circular cylinder x 2 y 2 2x. olution The surface lies in the plane 2x+3y+6z 6 so we use this to calculate d ifferentiating the equation for the plane with respect to x gives, thus, 1/3. ifferentiating the equation for the plane with respect to y gives, Hence, y 1 + ( )2 + ( y )2 thus, y 1/ / ( )2 + ( y )2 dxdy. Then the area of is found be calculating the suface integral over for the function f(x, y, z) 1. The the projection of the surface,, onto the x y-plane is given by {(x, y) : x 2 2x+y 2 (x 1) 2 + y 2 1}. Hence the area of is given by 1d dxdy 7 1dxdy Area of 7 6 π. Note, since is a cricle or radius 1 centred at (1, ) the area of is the area of a unit circle which is π. Example 5.8 Use Gauss ivergence Theorem to evaluate I x 4 y + y 2 z 2 + xz 2 d, where is the entire surface of the sphere x 2 + y 2 + z

5 olution In order to apply Gauss ( ivergence ) Theorem we first need to determine F and the unit normal n to the surface. The normal is f, f y, f (2x, 2y, 2z), where f(z, y, z) x 2 + y 2 + z 2 1. We require the unit normal, so n (2x, 2y, 2z)/ (2x, 2y, 2z) (2x, 2y, 2z)/2 (x, y, z). To find F (F 1, F 2, F 3 ) we note that F n x 4 y + y 2 z62 + xz 2 F 1 x + F 2 y + F 3 z Hence, comparing terms we have F 1 x 3 y, F 2 yz 2 and F 3 xz. Applying the ivergence Theorem noting that is the volume enclosed by the sphere gives Remarks I F nd div Fdxdydz 3x 2 y + z 2 + xdxdydz + z 2 dxdydz + 2π dφ π 1 π 2π cos 2 θ sin θ r 2 cos 2 θr 2 sin θdr 1 2π π As is a sphere it is natural to use spherical polar coordinates to solve the integral. Thus, x r cosφsin θ, y r sin φsin θ, and z r cosθ and dxdydz r 2 sin θ. 2. 3x2 ydxdydz and xdxdydz from the symmetry of the cosine and sine functions. We look at the signs in each quadrant as φ changes. Think about a fixed θ. cosφ and sinφ terms in x 2 y and x then have the following signs r 4 dr Quadrant Total cos φ + + sin φ + + x 2 y + + x + + The positive and negative contribution from the integral cancel out in these two cases so the integrals are zero. Example 5.9 Find I F n d where F (2x, 2y, 1) and where is the entire surface consisting of 1 the part of the paraboloid z 1 x 2 y 2 with z together with 2 disc {(x, y) : x 2 + y 2 1}. Here n is the outward pointing unit normal. 5

6 olution Applying the ivergence Theorem noting that is the volume enclosed by 1 and 2 and divf gives I F nd div Fdxdydz 4dxdydz {(x,y:)x 2 +y 2 1} {(x,y:)x 2 +y 2 1} 2π 1 (1 r 2 )r dr 4 2π(1/2 1/4) 2π. 1 x 2 y 2 dxdy 1dz 1 x 2 y 2 dxdy Example 5.1 ector fields and W are defined by (2x 3y + z, 3x y + 4z, 4y + z) W (2x 4y 5z, 4x + 2y, 5x + 6z). One of these is conservative while the other is not. etermine which is conservative and denote it by F. Find a potential function φ for F and evaluate F dr, where is the curve from A(1,,) to B(,,1) in which the plane x + z 1 cuts the hemisphere given by x 2 + y 2 + z 2 1, y. olution We have i j k curl y 2x 3y + z 3x y + 4z 4y + z (, 1, ). ince curl, F is NOT conservative. We have i j k curlw y 2x 4y 5z 4x + 2y 5x + 6z (,, ). 6

7 ince curl, F is conservative. uppose that grad φ W. Then φ 2x 4y 5z, (1) φ 4x + 2y, y (2) φ 5x + 6z. (3) Integrating (1) with respect to x, holding the other variables constant, we get φ dx x 2 4yx 5zx + A(y, z), y,z fixed2x 4y 5z where A is an arbitrary function. ubstituting this expression into (2) gives, 4x + A y 4x + 2y, A i.e. y 2y, and therefore where B is an arbitrary function, giving Finally, substituting this into (3) gives A(y, z) z fixed (2y)dy y 2 + B(z), φ x 2 4yx 5zx + y 2 + B(z). 5x + db dz 5x + 6z, db i.e. dz 6z, so that B 3z 2 +, where is a constant. Hence, by taking we obtain a potential φ x 2 4yx 5zx + y 2 x + 3z 2. Remark Notice that the potential function is not unique; we may always add an arbitrary constant to a potential and it remains a potential. o the line integral is: F dr div φ dr φ(,, 1) φ(1,, )