Field EvaluationLevel: 110 - 120(Requires 110) Patch PatchXP: 20,560Rep: +150 Talanji's ExpeditionRewards:Spare Stabilizer Bandor Nazmani Warplate Wargreavesor Loa Speaker's Trousersor Death-Pledged Legguardsor Ritual Flayer's Legwraps4680 1 Objectives 2 Description 3 Rewards 4. It will not play radio stations. The structure of the field is a simple anticline, 4 km long and 3 km wide, with a main east-west fault dividing the north and south areas (Fig. The geology is complex; the field comprises more than 500 hydrocarbon accumulations, stacked between 300 m to 4000 m subsea, trapped in channel-sand and sand-bar reservoirs deposited in a fluvio.
Show Mobile NoticeShow All NotesHide All NotesYou appear to be on a device with a 'narrow' screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 5-4 : Line Integrals of Vector Fields
In the previous two sections we looked at line integrals of functions. In this section we are going to evaluate line integrals of vector fields. We’ll start with the vector field,
[vec Fleft( {x,y,z} right) = Pleft( {x,y,z} right)vec i + Qleft( {x,y,z} right)vec j + Rleft( {x,y,z} right)vec k] and the three-dimensional, smooth curve given by
[vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec khspace{0.25in}hspace{0.25in}a le t le b] The line integral of (vec F) along (C) is
[intlimits_{C}{{vec Fcenterdot d,vec r}} = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}] Note the notation in the integral on the left side. That really is a dot product of the vector field and the differential really is a vector. Also, (vec Fleft( {vec rleft( t right)} right)) is a shorthand for,
[vec Fleft( {vec rleft( t right)} right) = vec Fleft( {xleft( t right),yleft( t right),zleft( t right)} right)] We can also write line integrals of vector fields as a line integral with respect to arc length as follows,
[intlimits_{C}{{vec Fcenterdot d,vec r}} = intlimits_{C}{{vec Fcenterdot vec T,ds}}] where (vec Tleft( t right)) is the unit tangent vector and is given by,
[vec Tleft( t right) = frac{{vec r'left( t right)}}{{left| {vec r'left( t right)} right|}}] 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.
[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = intlimits_{C}{{vec Fcenterdot vec T,ds}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot frac{{vec r'left( t right)}}{{left| {vec r'left( t right)} right|}},left| {vec r'left( t right)} right|,dt}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}end{align*}] In general, we use the first form to compute these line integral as it is usually much easier to use. Let’s take a look at a couple of examples.
Example 1 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = 8{x^2}y,z,vec i + 5z,vec j - 4x,y,vec k) and (C) is the curve given by (vec rleft( t right) = t,vec i + {t^2},vec j + {t^3},vec k), (0 le t le 1). Show SolutionOkay, we first need the vector field evaluated along the curve.
[vec Fleft( {vec rleft( t right)} right) = 8{t^2}left( {{t^2}} right)left( {{t^3}} right)vec i + 5{t^3},vec j - 4tleft( {{t^2}} right)vec k = 8{t^7},vec i + 5{t^3},vec j - 4{t^3},vec k] Next, we need the derivative of the parameterization.
[vec r'left( t right) = ,vec i + 2t,vec j + 3{t^2},vec k] Field Evaluation Ul
Finally, let’s get the dot product taken care of.
[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 8{t^7} + 10{t^4} - 12{t^5}] The line integral is then,
[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,0}}^{{,1}}{{8{t^7} + 10{t^4} - 12{t^5},dt}} & = left. {left( {{t^8} + 2{t^5} - 2{t^6}} right)} right|_0^1 & = 1end{align*}] Example 2 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = x,z,vec i - y,z,vec k) and (C) is the line segment from (left( { - 1,2,0} right)) to (left( {3,0,1} right)). Show SolutionWe’ll first need the parameterization of the line segment. We saw how to get the parameterization of line segments in the first section on line integrals. We’ve been using the two dimensional version of this over the last couple of sections. Here is the parameterization for the line.
[begin{align*}vec rleft( t right) & = left( {1 - t} right)leftlangle { - 1,2,0} rightrangle + tleftlangle {3,0,1} rightrangle & = leftlangle {4t - 1,2 - 2t,t} rightrangle ,hspace{1.0in} 0 le t le 1end{align*}] So, let’s get the vector field evaluated along the curve.
[begin{align*}vec Fleft( {vec rleft( t right)} right) & = left( {4t - 1} right)left( t right),vec i - left( {2 - 2t} right)left( t right),vec k & = left( {4{t^2} - t} right)vec i - left( {2t - 2{t^2}} right)vec kend{align*}] ![Evaluation Evaluation](/uploads/1/1/8/3/118355132/781164737.jpg)
Field 3 Evaluation Definition
Now we need the derivative of the parameterization.
[vec r'left( t right) = leftlangle {4, - 2,1} rightrangle ] The dot product is then,
[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 4left( {4{t^2} - t} right) - left( {2t - 2{t^2}} right) = 18{t^2} - 6t] The line integral becomes,
[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}}& = int_{{,0}}^{{,1}}{{18{t^2} - 6t,dt}} & = left. {left( {6{t^3} - 3{t^2}} right)} right|_0^1 & = 3end{align*}] Let’s close this section out by doing one of these in general to get a nice relationship between line integrals of vector fields and line integrals with respect to (x), (y), and (z).
Given the vector field (vec Fleft( {x,y,z} right) = P,vec i + Q,vec j + R,vec k) and the curve (C) parameterized by (vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec k), (a le t le b) the line integral is,
[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,a}}^{{,b}}{{left( {P,vec i + Q,vec j + R,vec k} right)centerdot left( {x',vec i + y',vec j + z',vec k} right),dt}} & = int_{{,a}}^{{,b}}{{Px' + Qy' + Rz',dt}} & = int_{{,a}}^{{,b}}{{Px',dt}} + int_{{,a}}^{{,b}}{{Qy',dt}} + int_{{,a}}^{{,b}}{{Rz',dt}} & = intlimits_{C}{{P,dx}} + intlimits_{C}{{Q,dy}} + intlimits_{C}{{R,dz}} & = intlimits_{C}{{P,dx}} + Q,dy + R,dzend{align*}] So, we see that,
[intlimits_{C}{{vec Fcenterdot d,vec r}} = intlimits_{C}{{P,dx}} + Q,dy + R,dz]
Note that this gives us another method for evaluating line integrals of vector fields.
This also allows us to say the following about reversing the direction of the path with line integrals of vector fields.
Fact
[intlimits_{{ - C}}{{vec Fcenterdot d,vec r}} = - intlimits_{C}{{vec Fcenterdot d,vec r}}]
This should make some sense given that we know that this is true for line integrals with respect to (x), (y), and/or (z) and that line integrals of vector fields can be defined in terms of line integrals with respect to (x), (y), and (z).