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) =

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.5: Polar Coordinates
Problem 97E
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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) and
Surface: parabolid z = 4 -x2 - y2
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.
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 below
Vector Field: F (x,y) = (z. - 3xy, 8>
and
Surface: paraboloid z = 4 - x - y2
Find the "curl" of the vector field.
of
dz dz
ду
A 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 formulas
we can see that the normal vector for a cone is
i- (2x, 2y. - 1)
Use the value for curl from the previous question and the normal vector to find their dot product.
[vxF]-ñ
A [vxF]. = 0
® [v xF].i = 5y
© [vxF]. i =
- 6
O (v xF]- n = -– 6xy
Given the vector field and surface below
Vector 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. Please
note that if your answer to the previous question does not contain any "z's", you may not have to plug anything into your previous
answer.
A v xF. i = 0
B v x F- n = 5y
© [v x F].= - 6xy
- &r
O [v xF]- =
9-
Given the vector field and surface below
Vector Field: F(x,y) = (z. - 3xy, 8) and
Surface: paraboloid z = 4 - x2 - y2
Now that we have taken care of the inside of our integral, we need to set up the double integral. Remember that this double integral
handles 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 am
arbitrarily going to set the height of the cone to "z= 3". Plugging this into our equation for the cone, we get
9 = 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 dy
O /* S [[v x F]. a] ndr de
Now that you have found the curl, the normal vector, the dot product, and set up the double integral, we can use Stoke's Theorem
to find the circulation for this situation.
Circulation =
@ / [[vxF] . ] as = 0
© [ [[vx#]- ï] as - - 15
© [[vxF]- #] as =-
- 4x
Transcribed Image Text:Vector Field: F(x,y.z) = (z. - 3xy, 8) and Surface: parabolid z = 4 -x2 - y2 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. 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 below Vector Field: F (x,y) = (z. - 3xy, 8> and Surface: paraboloid z = 4 - x - y2 Find the "curl" of the vector field. of dz dz ду A 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 formulas we can see that the normal vector for a cone is i- (2x, 2y. - 1) Use the value for curl from the previous question and the normal vector to find their dot product. [vxF]-ñ A [vxF]. = 0 ® [v xF].i = 5y © [vxF]. i = - 6 O (v xF]- n = -– 6xy Given the vector field and surface below Vector 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. Please note that if your answer to the previous question does not contain any "z's", you may not have to plug anything into your previous answer. A v xF. i = 0 B v x F- n = 5y © [v x F].= - 6xy - &r O [v xF]- = 9- Given the vector field and surface below Vector Field: F(x,y) = (z. - 3xy, 8) and Surface: paraboloid z = 4 - x2 - y2 Now that we have taken care of the inside of our integral, we need to set up the double integral. Remember that this double integral handles 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 am arbitrarily going to set the height of the cone to "z= 3". Plugging this into our equation for the cone, we get 9 = 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 dy O /* S [[v x F]. a] ndr de Now that you have found the curl, the normal vector, the dot product, and set up the double integral, we can use Stoke's Theorem to find the circulation for this situation. Circulation = @ / [[vxF] . ] as = 0 © [ [[vx#]- ï] as - - 15 © [[vxF]- #] as =- - 4x
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