Introduction to Surface integrals of vector fields over parameterized surface:
The surface integral of vector field F on R3 over a surface S described in parametric form by r : D ? R3 is given by the number :
?? S F`*` dS = ?? DF(r(u,v)) `*` (ru `xx` rv) du dv
Surface integrals of vector fields for graphs:
Consider the vector field F = Pi + Qj + Rk and the surface S given by the graph z = f(x, y) on a domain D. Then the surface integral of vector field F over the surface S is :
`intint` S F`*` dS = ?? S F`*` ndS = ?? D(-Pfx - Qfy + R)dx dy
Example for Surface Integrals of Vector Fields
2? ?
Let S be the unit sphere given by the parameterization:
x = cos ? sin f , y = sin ? sin f , z = cos f for 0 = ? = 2p , 0 = f = p
Find the surface integral of vector field F(x, y, z) = xi + yj + zk over the surface S.
Solution:
Here we have r(?, f) = ( cos ? sin f, sin ? sin f, cos f)
Hence,
r? = (-sin ? sin f, cos ? sin f, 0)
rf = (cos ? cos f, sin ? cos f, -sin f)
r? × rf = (-sin2 f cos ?, -sin2 f sin ?, -sin f cos f)
Next we evaluate,
F · (r? × rf) = -sin f
Hence,
?? S F· dS = ?? DF(r(?, f)) · (r? × rf) d ? d ?
= $\int_{0}^{2\Pi }\int_{0}^{\Pi } -sin \varphi d\varphi d\theta$
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Exercise on Surface Integrals of Vector Fields
Consider the vector field F = x2i + y2j + zk. Let S be the surface given by the graph of the function z = x + y + 1 over the rectangle 0=x=1 , 0=y=1. Evaluate ?? S F`*` ndS
Hint: Here z = f(x, y) = x+y+1. Hence find fx and fy and plug in the formula given above for surface integrals of vector fields for graphs.
Conclusion on Surface Integrals of Vector Fields:
Surface integrals of vector fields are also referred to as flux across S. In other words, it is the volume of fluid moving with a velocity described by the vector field F(x,y,z) crossing the surface S per unit time.
The surface integral of vector field F on R3 over a surface S described in parametric form by r : D ? R3 is given by the number :
?? S F`*` dS = ?? DF(r(u,v)) `*` (ru `xx` rv) du dv
Surface integrals of vector fields for graphs:
Consider the vector field F = Pi + Qj + Rk and the surface S given by the graph z = f(x, y) on a domain D. Then the surface integral of vector field F over the surface S is :
`intint` S F`*` dS = ?? S F`*` ndS = ?? D(-Pfx - Qfy + R)dx dy
Example for Surface Integrals of Vector Fields
2? ?
Let S be the unit sphere given by the parameterization:
x = cos ? sin f , y = sin ? sin f , z = cos f for 0 = ? = 2p , 0 = f = p
Find the surface integral of vector field F(x, y, z) = xi + yj + zk over the surface S.
Solution:
Here we have r(?, f) = ( cos ? sin f, sin ? sin f, cos f)
Hence,
r? = (-sin ? sin f, cos ? sin f, 0)
rf = (cos ? cos f, sin ? cos f, -sin f)
r? × rf = (-sin2 f cos ?, -sin2 f sin ?, -sin f cos f)
Next we evaluate,
F · (r? × rf) = -sin f
Hence,
?? S F· dS = ?? DF(r(?, f)) · (r? × rf) d ? d ?
= $\int_{0}^{2\Pi }\int_{0}^{\Pi } -sin \varphi d\varphi d\theta$
Between, if you have problem on these topics how to multiply 3 fractions, please browse expert math related websites for more help on how to divide and multiply fractions.
Exercise on Surface Integrals of Vector Fields
Consider the vector field F = x2i + y2j + zk. Let S be the surface given by the graph of the function z = x + y + 1 over the rectangle 0=x=1 , 0=y=1. Evaluate ?? S F`*` ndS
Hint: Here z = f(x, y) = x+y+1. Hence find fx and fy and plug in the formula given above for surface integrals of vector fields for graphs.
Conclusion on Surface Integrals of Vector Fields:
Surface integrals of vector fields are also referred to as flux across S. In other words, it is the volume of fluid moving with a velocity described by the vector field F(x,y,z) crossing the surface S per unit time.
