Let S be the surface of the cone z=√x² y² bounded by the planes z=0 and z=3. Further let C be the closed surface forming the boundary of the surface S. A vector field is such that curl F is -xi-yj. Find absolute value of line integral F.dr?

Question

Answer

Given information:
S is the surface of the cone z=√(x² + y²) bounded by the planes z=0 and z=3.
C is the closed surface forming the boundary of S.
The vector field F has curl F = -xi - yj.

To find:
The absolute value of the line integral F.dr.

Explanation:
To find the line integral of a vector field F along a curve C, we can use the line integral formula:

∫(C) F.dr = ∫(a to b) F(r(t)).r'(t)dt

Where r(t) is the parameterization of the curve C, and r'(t) is the derivative of r(t) with respect to t.

Parameterization of C:
Since C forms the boundary of S, we can parameterize C using the equation of the cone surface:

r(t) = (t, t, √(t² + t²)) = (t, t, √2t²) = (t, t, √2)t

Where t ranges from 0 to 2π.

Calculating F(r(t)):
To calculate F(r(t)), we substitute the values of r(t) into the vector field F:

F(r(t)) = -(t)i - (t)j

Calculating r'(t):
r'(t) = (1, 1, √2)

Calculating F.dr:
Now we can calculate F.dr using the line integral formula:

∫(C) F.dr = ∫(0 to 2π) F(r(t)).r'(t)dt
= ∫(0 to 2π) [-(t)i - (t)j].(1, 1, √2) dt
= ∫(0 to 2π) [-t - t√2] dt
= -∫(0 to 2π) t(1 + √2) dt
= -(1 + √2) ∫(0 to 2π) t dt
= -(1 + √2) [t²/2] (0 to 2π)
= -(1 + √2) [(4π² - 0)/2]
= -2π²(1 + √2)

Calculating the absolute value:
Since the question asks for the absolute value of the line integral, we take the magnitude:

|F.dr| = |-2π²(1 + √2)|
= 2π²(1 + √2)

Therefore, the absolute value of the line integral F.dr is 2π²(1 + √2).

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