For Questions 7, 8, 9, 10, and 11 we are given a vector field and a closed path. Vector Field: F(x,y.z) = (z. - 3xy, 8) andSurface: parabolid z = 4 -x2 - y2Since we do not have our path as a vector, we can use Stoke's Theorem to find the Circulation.Circulation =The "n" refers to the normal vector of the surface. Determining the normal vector is the most difficult part of using Stoke's Theorem.Notice how the path is not present at all inside the integral. The double integral, itself, will handle the path that our object is taking.Given the vector field and surface belowVector Field: F (x,y) = (z. - 3xy, 8>andSurface: paraboloid z = 4 - x - y2Find the "curl" of the vector field.ofdz dzдуA vXF = (0. - 3x, 0)B v XF = (1. - 3, 8)© vxF = (0. 1, - 3y)O vxF = (0, 0, 0>Our surface formula is for a paraboloid. Using our formula sheet for Parameterized and Explicit Surface formulaswe can see that the normal vector for a cone isi- (2x, 2y. - 1)Use the value for curl from the previous question and the normal vector to find their dot product.[vxF]-ñA [vxF]. = 0® [v xF].i = 5y© [vxF]. i =- 6O (v xF]- n = -– 6xyGiven the vector field and surface belowVector Field: F(x,y) = (2. - 3xy, 8> and Surface: paraboloid z = 4 - x2 - y?Take the equation of your surface and solve for "z". Once you do that, plug it into your answer from the previous question. Pleasenote that if your answer to the previous question does not contain any "z's", you may not have to plug anything into your previousanswer.A v xF. i = 0B v x F- n = 5y© [v x F].= - 6xy- &rO [v xF]- =9-Given the vector field and surface belowVector Field: F(x,y) = (z. - 3xy, 8) andSurface: paraboloid z = 4 - x2 - y2Now that we have taken care of the inside of our integral, we need to set up the double integral. Remember that this double integralhandles the path that our object is taking. Recall that our surface is a cone, and so our path around this is a circle.Since our surface is a cone, we need to determine the height of the cone so that we can determine the size of our circular path. I amarbitrarily going to set the height of the cone to "z= 3". Plugging this into our equation for the cone, we get9 = x2 + y?as the equation for our path. With all of this in mind, let's set up the double integral for our path.® /T [[v xF]. #) dx dy© S S[[v x#]. #] de dyO /* S [[v x F]. a] ndr deNow that you have found the curl, the normal vector, the dot product, and set up the double integral, we can use Stoke's Theoremto find the circulation for this situation.Circulation =@ / [[vxF] . ] as = 0© [ [[vx#]- ï] as - - 15© [[vxF]- #] as =-- 4x

Answered: Image /qna-images/answer/51ed12f8-c06e-453e-9e53-3a3874ead9cc.jpg

since we do not have our path as a vector, we can use Stoke's Theorem to find the Circulation. Circulation = The "n" refers to the normal vector of the surface. Determining the normal vector is the most difficult part of using Stoke's Theorem. lotice how the path is not present at all inside the integral. The double integral, itself, will handle the path that our object is taking iven the vector field and surface below "ector Field: F(x,y) =

since we do not have our path as a vector, we can use Stoke's Theorem to find the Circulation. Circulation = The "n" refers to the normal vector of the surface. Determining the normal vector is the most difficult part of using Stoke's Theorem. lotice how the path is not present at all inside the integral. The double integral, itself, will handle the path that our object is taking iven the vector field and surface below "ector Field: F(x,y) =

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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