Solutions to Vector Calculus Problems

Solution
Q1)
F(x, y, z) = (2xz + zsiny, xzcosy, x2 + xsiny)
Curl F =
| ^i ^j ^k
∂
∂ x
∂
∂ y
∂
∂ z
∂ xz + zsiny xzcosy x2 + xsiny
|
= (xcosy – xcosy) ^i−¿
= 0 ^i−0 ^j+ 0 ^k
= 0
F is conservative in R3
To determine a scalar potential function
∂ f
∂ y =¿2xz + zsiny,
∂ f
∂ y =xzcosy ,
∂ f
∂ y =x2 + xsiny
Integrating ∂ f
∂ y =¿2xz + zsiny
f = x2z = xysiny + g(y,z)
The partial derivative of f = x2z = xysiny + g(y,z) with respect to y
∂ f
∂ y =¿xycosy + ∂ g ( y , z )
∂ y
Comparing to ∂ f
∂ y =xzcosy
xzcosy + ∂ y
∂ y = xzcosy
∂ y
∂ y = 0

g(y, z) = h(z)
Partial derivative of f = x2z = xysiny + g(y,z) with respect to z
∂ f
∂ z =x2 + xsiny+ h '(z )
Comparing this to ∂ f
∂ y =x2 + xsiny
x2+ xsiny+h' ( z ) =x2 + xsiny
= h’(z) = 0
= h(z) = C where C is a constant.
f(x, y, z) = x2z + xzsiny + C
By fundamental theorem of line integral
∫
C
❑
F . dr=f (−1 , 0 , 4 )−f (1 , 0 ,0)
f(-1, 0, 4) = 4 + C
f(1, 0, 0) = 0 + C
then f(-1, 0, 4) – f(1, 0, 0) = 4
therefore,
∫
C
❑
F . dr=4
Q2)
Using stokes theorem
∫∫
S
❑
( ∇∗F )∗nds=∫
C
❑
F . dr
The parabolic z = 4 – x2 – y2 intersects xy- plane as a circle x2 + y2 = 4
r(t) = (2cost, 2sint, 0), 0 > t ≤ 2 π

r’(t) = (-2sint, 2cost, 0)
∫
C
❑
F . dr=∫
t
❑
F ( r ( t ) )∗r' ( t ) dt
= ∫
t
❑
(2 sin t ¿ e0−2 sin t¿ ,2 cos t∗e0 , 1+ 4 sin t cos t e0 )(−2 sint , 2cos t , 0)∗dt ¿
= ∫
t
❑
(0,2 cos t ¿,1+ 4 sin t cos t) (−2sin t ,2 cos t , 0 ) dt ¿
= ∫
t
❑
4 cos2 t dt=4∫
t
❑ 1+cos 2 t
2 dt
= 2[(t)02π + ( sin 2t
2 )02π]
= 2[2 π + 0 ¿
= 4 π
= ∫∫
S
❑
( ∇∗F ) ∗nds=4 π
Q3)
By Gauss divergence theorem
∫∫
S
❑
F . ^n ds=∫∫∫
V
❑
¿⃗ F dv, where V is the volume bounded by sphere S
divF = ∇∗F= ∂
∂ x ( x ) + ∂
∂ y ( 2 y ) + ∂
∂ z (x z2)
= 1 + 2 + 2xz
= 3 + 2xz
∫∫
S
❑
F . ^n ds=∫∫∫
V
❑
( 3+2 x z ) dxdydz ................eq1
Converting this to spherical polar coordinate system, then