Use Stokes' Theorem to evaluate
C
F · dr
where C is oriented counterclockwise as viewed from above.
F(x, y, z) = yzi + 9xzj + exyk,
C is the circle
x2 + y2 = 1, z = 3.
To evaluate this integral using Stokes' Theorem, we need to follow these steps:
Step 1: Determine the surface bounded by the given curve.
The given curve C is a circle centered at the origin and lying on the xy-plane. Since z = 3, the curve is a circle in the plane z = 3. Therefore, the surface bounded by this curve is the portion of the plane z = 3 that lies within the circle.
Step 2: Compute the curl of the vector field F.
The curl of a vector field F(x, y, z) = yzi + 9xzj + exyk is defined as:
curl(F) = ∇ × F
= (∂Fz/∂y - ∂Fy/∂z)i + (∂Fx/∂z - ∂Fz/∂x)j + (∂Fy/∂x - ∂Fx/∂y)k
Calculating the partial derivatives:
∂Fz/∂y = z = 0
∂Fy/∂z = 0
∂Fx/∂z = 9x
∂Fz/∂x = 0
∂Fy/∂x = 0
∂Fx/∂y = z = 0
Therefore, the curl of F is:
curl(F) = 0i + 9xj + 0k
= 9xj
Step 3: Find the unit normal vector to the surface.
Since the surface lies on the plane z = 3, the normal vector to the plane is k. So, the unit normal vector to the surface is:
n = k/|k| = k/√(1^2 + 1^2)
= k/√2
Step 4: Calculate the flux integral using Stokes' Theorem.
According to Stokes' Theorem, the flux integral of F across the surface bounded by C is equal to the line integral of F along the curve C. Therefore, we have:
∬S curl(F) · dS = ∮C F · dr
Since the given curve C is a circle in the plane z = 3, we can parameterize it as:
r(t) = cos(t)i + sin(t)j for t in [0, 2π].
Now we can calculate the line integral:
∮C F · dr = ∫[0, 2π] F(r(t)) · r'(t) dt
Plugging in the values of F and r(t):
∮C F · dr = ∫[0, 2π] (sin(t)*cos(t)i + 9cos(t)*sin(t)j + e*sin(t)*cos(t)k) · (-sin(t)i + cos(t)j) dt
Simplifying the expression:
∮C F · dr = ∫[0, 2π] (-sin^2(t) + 9cos^2(t) + e*sin(t)*cos(t)) dt
Now, you can evaluate this integral to find the final result.
To use Stokes' Theorem to evaluate the line integral, first, we need to find the curl of the vector field.
Given vector field F(x, y, z) = yzi + 9xzj + exyk, we can find the curl using the formula:
curl(F) = ∇ x F
where ∇ is the del operator.
The del operator in Cartesian coordinates is given by:
∇ = (∂/∂x)i + (∂/∂y)j + (∂/∂z)k
Now, we need to find the components of the curl of F:
curl(F) = (∂Fz/∂y - ∂Fy/∂z)i + (∂Fx/∂z - ∂Fz/∂x)j + (∂Fy/∂x - ∂Fx/∂y)k
Let's calculate the partial derivatives of F:
∂Fx/∂x = 0
∂Fx/∂y = z
∂Fx/∂z = 9x
∂Fy/∂x = -z
∂Fy/∂y = 0
∂Fy/∂z = 0
∂Fz/∂x = 0
∂Fz/∂y = 0
∂Fz/∂z = exy
Now, we can plug these values into the curl formula:
curl(F) = (0 - 0)i + (0 - (-z))j + (0 - 9x)k
= zi + zj - 9xk
Next, we need to find the surface area that C encloses. C is a circle with the equation x^2 + y^2 = 1 and z = 3. This represents a circle in the xy-plane centered at the origin with radius 1.
To parametrize the circle, we can use the parameter t:
x = cos(t)
y = sin(t)
where t ranges from 0 to 2π.
Now, we can use Stokes' Theorem to compute the line integral:
∫∫(curl(F) · dS) = ∫∫(zi + zj - 9xk) · (dA)
Since C is a circle in the xy-plane, the normal vector to the surface will be k, and dA = dx dy.
Therefore, the line integral simplifies to:
∫∫(zi + zj - 9xk) · (k) dx dy
Taking only the k-component:
∫∫(-9x) dx dy
Now, we need to find the limits of integration for x and y. Since C is a circle with radius 1 centered at the origin, we can integrate over the range 0 to 2π for t, which corresponds to x and y ranging from -1 to 1.
Therefore, the final integral becomes:
∫∫(-9x) dx dy, where x and y range from -1 to 1.
Evaluating the integral:
∫∫(-9x) dx dy = -9 ∫∫(x) dx dy
Since x is an odd function, the integral of an odd function over a symmetric region is zero.
Therefore, the line integral is equal to zero.
So, the answer is 0.
curl of F over the surface
curl is
i j k
d/dx d/dy d/dz
fx fy fz
is
=(dFz/dy-dFy/dz)i
+(dFx/dz-dFz/dx)j
+(dFy/dx-dFx/dy)k
here Curl F
=(ex-9x)i
+(y-ey)j
+(9z-z)k