nce we do not have our path as a vector, we can use Stoke's Theorem to find th nd the "curl" of the vector field. og of XF - dy A vxi = (2. - 3, s) 8 vxF = (0. 2, - 3) © vxF = (-2, 5, -3) O vxi - (0. 0, 0) e can see that the normal vector for a parabolaid is - (2x, 2y, – 1) se the value for curl from the previous question and the normal vector to find s A [v xF] -ñ = o ® [vxF] . = 4 - 6xy - 8

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Chapter1: Functions And Models
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We are given a vector field and a closed path.

Vector Field: F(x,y.z) = (2:, - 3x, 8) and Surface: paraboloid z = 16 - x2 - y2 and Path: x² + y? = 16
Since we do not have our path as a vector, we can use Stoke's Theorem to find the Circulation.
Find the "curl" of the vector field.
og of
VxF -
dy
dz dz
dx
A v XF = (2, - 3, 8)
® vXF = (0, 2. - 3)
© vxF = (-2, 5, - 3)
O vxF = (0, 0, o>
we can see that the normal vector for a paraboloid is
n = (2x, 2y, - 1>
Use the value for curl from the previous question and the normal vector to find their dot product.
[vxF]-
A (vxF- = 0
B V xF - n = 4xz - 6xy - 8
© [v xF- i = 4x - 3y + 8
O (v xF- = 4y + 3
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 [vxF]. n = 4y + 3
® [vxF] -n = 0
© vxF. n = 64x - 4x - 4xy² – 6xy – 8
D v xF. = 4x – 3y + 8
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 paraboloid, and so our path around this is a circle that is
placed at the origin.
Since our path is a circle placed at the origin, we can let the "z" component of our surface equation be "0". This gives us an equation
for our path of
24 y2 - 16
With all of this in mind, let's set up the double integral for our path.
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.
- llvxi]. #] as
Circulation =
@ I [[vxF] . ] as = 0
® / [[vxF] . ] as = 881
656
O [ [[vxF] . #] as = 4tx
Transcribed Image Text:Vector Field: F(x,y.z) = (2:, - 3x, 8) and Surface: paraboloid z = 16 - x2 - y2 and Path: x² + y? = 16 Since we do not have our path as a vector, we can use Stoke's Theorem to find the Circulation. Find the "curl" of the vector field. og of VxF - dy dz dz dx A v XF = (2, - 3, 8) ® vXF = (0, 2. - 3) © vxF = (-2, 5, - 3) O vxF = (0, 0, o> we can see that the normal vector for a paraboloid is n = (2x, 2y, - 1> Use the value for curl from the previous question and the normal vector to find their dot product. [vxF]- A (vxF- = 0 B V xF - n = 4xz - 6xy - 8 © [v xF- i = 4x - 3y + 8 O (v xF- = 4y + 3 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 [vxF]. n = 4y + 3 ® [vxF] -n = 0 © vxF. n = 64x - 4x - 4xy² – 6xy – 8 D v xF. = 4x – 3y + 8 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 paraboloid, and so our path around this is a circle that is placed at the origin. Since our path is a circle placed at the origin, we can let the "z" component of our surface equation be "0". This gives us an equation for our path of 24 y2 - 16 With all of this in mind, let's set up the double integral for our path. 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. - llvxi]. #] as Circulation = @ I [[vxF] . ] as = 0 ® / [[vxF] . ] as = 881 656 O [ [[vxF] . #] as = 4tx
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