7. Use Stokes Theorem to calculate the circulation of the vector field F =(2²+z)i + + y² + ² = 1 with (y² + 2x)j + (2²-y)k along the curve of intersection of the sphere the cone z = √² + y² traversed in the counterclockwise direction around the z-axiz when viewed from above.
7. Use Stokes Theorem to calculate the circulation of the vector field F =(2²+z)i + + y² + ² = 1 with (y² + 2x)j + (2²-y)k along the curve of intersection of the sphere the cone z = √² + y² traversed in the counterclockwise direction around the z-axiz when viewed from above.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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is this solved correctly?

Transcribed Image Text:7.
Use Stokes Theorem to calculate the circulation of the vector field F =(2²+z)i +
(y² + 2x)j + (2²-y)k along the curve of intersection of the sphere a + y² + ² = 1 with
the cone z = √² + y² traversed in the counterclockwise direction around the z-axiz when
viewed from above.

Transcribed Image Text:Stokes = $uxFondo = √ FIRCED OF OF
(t)=< cost, Sint, 07
a? <-sint, cost, or
●
at
Fir(t) = <cos²t, sin²+ + 2 cost, -sint> · <-sint, cost, os
J² >cos²€ sint + Sin²t cost + 2c06²+ dt
기
U= cost
u= Sint
du sint
du= Cost
2
=f}" u²du
1)
U² =
2T
2 2 cos? tdt
= 2
Z
}(1-1)=0
21" 1 + cos² dt
2
2πT
= 25th = / + cos20
10
= 2
N
t
2 ( ½ + + Sin²²) | 20
花
4
10
2 (240) 211
-sin2t 2
N
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