Using Stokes' theorem evaluate ∫c F.dl where F = yi^+ xz^3 j^-z y^3 k^ and C is the circle x^2+ y^2 = 4; z = -3.
To evaluate the line integral ∫c F·dl using Stokes' theorem, we need to calculate the curl of the vector field F and then find the surface integral of the curl over the surface bounded by the given curve C.
1. Find the curl of the vector field F:
The curl of a vector field F = P i + Q j + R k is given by the determinant of the following expression:
∇ x F = ( ∂R/∂y - ∂Q/∂z ) i + ( ∂P/∂z - ∂R/∂x ) j + ( ∂Q/∂x - ∂P/∂y ) k
For our vector field F = y i + xz^3 j - zy^3 k, we have:
P = y, Q = xz^3, R = -zy^3
Taking the partial derivatives, we get:
∂P/∂z = 0, ∂Q/∂x = z^3, ∂R/∂y = -3zy^2
∂Q/∂z = 3xz^2, ∂R/∂x = 0, ∂P/∂y = 1
Therefore, the curl of F, denoted as ∇ x F, is given by:
∇ x F = (0 - 3zy^2) i + (z^3 - 0) j + (x - 1) k
= -3zy^2 i + z^3 j + (x - 1) k
2. Parametrize the surface bounded by the given curve C:
The surface bounded by the circle x^2 + y^2 = 4 and z = -3 can be parametrized as:
ϕ(u, v) = (2 cos(u), 2 sin(u), -3), where 0 ≤ u ≤ 2π
3. Calculate the surface integral:
∫∫S (∇ x F) · dS
Using the parametrization ϕ(u, v), we can calculate the partial derivatives of ϕ with respect to u and v:
∂ϕ/∂u = (-2 sin(u), 2 cos(u), 0)
∂ϕ/∂v = (0, 0, 0)
Since ∂ϕ/∂v is zero, the surface integral simplifies to:
∫∫S (∇ x F) · dS = ∫∫S (∇ x F) · (∂ϕ/∂u x ∂ϕ/∂v) dA
The cross product of the partial derivatives is:
(∂ϕ/∂u x ∂ϕ/∂v) = (2 cos(u), 2 sin(u), 0)
Now we can calculate the surface integral:
∫∫S (∇ x F) · (∂ϕ/∂u x ∂ϕ/∂v) dA = ∫∫S (-3zy^2, z^3, x -1) · (2 cos(u), 2 sin(u), 0) dA
Since z is constant (-3) and y = 2 sin(u), the result is:
∫∫S (-6 sin(u)^2, (-3)^3, 2 cos(u) - 1) · (2 cos(u), 2 sin(u), 0) dA
Simplifying further:
∫∫S (-6 sin(u)^2, -27, 2 cos(u) - 1) · (2 cos(u), 2 sin(u), 0) dA
= ∫∫S (-12 sin(u)^2 cos(u), -54 + 0, 4 cos^2(u) - 2 cos(u) - 2 sin(u)) dA
Now, we need to calculate the area element dA. Since the surface is a circle of radius 2, dA can be replaced by the area of a circle of radius 2:
dA = π(2)^2 = 4π
Substituting in the values:
= ∫∫S (-12 sin(u)^2 cos(u), -54, 4 cos^2(u) - 2 cos(u) - 2 sin(u)) 4π
Now, we integrate with respect to u:
= 4π ∫[0 to 2π] (-12 sin(u)^2 cos(u), -54, 4 cos^2(u) - 2 cos(u) - 2 sin(u)) du
Evaluating the integral term by term, we get:
= 4π [0, -54u, 4(u + sin(u)cos(u) - u - sin(u))] evaluated from 0 to 2π
Simplifying further, we get:
= 4π [0, -108π, 0]
Therefore, the value of the line integral ∫c F·dl is -108π.