(a)Find the work done by the field
F= yi+zj+xk in travelling around the curve X^2+y^2=4, z=0.
(b) state stokes theorem and verify it for the field F given in (a), over the disc x^2+y^2<4, z=0
To find the work done by the field F = yi + zj + xk in traveling around the curve X^2 + y^2 = 4, z = 0, we can use the line integral formula for work:
Work = ∮ F · dr
Where F is the field vector and dr is the differential displacement vector along the curve.
To compute this integral, we need to parameterize the curve. Since the curve is given by X^2 + y^2 = 4 and z = 0, we can use polar coordinates to parameterize it. Let's express x and y in terms of θ:
x = 2cosθ,
y = 2sinθ,
z = 0.
Now, let's find the differential displacement vector dr. Since x = 2cosθ, y = 2sinθ, and z = 0, we have:
dr = dx i + dy j + dz k
= (-2sinθ) dθ i + (2cosθ) dθ j + 0 k
= (-2sinθ)i dθ + (2cosθ)j dθ.
Next, let's calculate F · dr:
F · dr = (yi + zj + xk) · ((-2sinθ)i dθ + (2cosθ)j dθ)
= (-2ysinθ)dθ + (2xcosθ)dθ
= (-2(2sinθ)sinθ)dθ + (2(2cosθ)cosθ)dθ
= -4sin^2θ dθ + 4cos^2θ dθ.
Now, let's evaluate the line integral:
Work = ∮ F · dr
= ∫[-π,π] (-4sin^2θ dθ + 4cos^2θ dθ).
Integrating with respect to θ, we get:
Work = [-4(θ + sin(2θ)/2) + 4(θ + sin(2θ)/2)] | [-π,π]
= -8π.
Therefore, the work done by the field F = yi + zj + xk in traveling around the curve X^2 + y^2 = 4, z = 0 is -8π.
Moving on to part (b) of the question:
Stokes' theorem relates the surface integral of a vector field over a surface S to the line integral of the vector field around the boundary curve C of that surface. Mathematically, it can be stated as follows:
∮(∇ × F) · dS = ∮F · dr,
where ∇ × F is the curl of the vector field F, dS is the vector representing the differential area element on the surface S, and dr is the differential displacement vector along the boundary curve C.
To verify Stokes' theorem for the field F = yi + zj + xk over the disk x^2 + y^2 < 4, z = 0, we need to calculate both sides of the equation.
First, let's find the curl of F:
∇ × F = (d/dx)i + (d/dy)j + (d/dz)k
= 0i + 0j + 1k.
Next, we need to find the outward pointing normal vector dS for the surface. Since the surface is given by x^2 + y^2 < 4, z = 0, the normal vector points in the positive z-direction. So, dS = k dA, where dA is the differential area element on the disk.
Since the field F is constant in the z-direction, F · dS = (yi + zj + xk) · (k dA) = (xk) · (k dA) = x dA.
Now, let's calculate the surface integral:
∮(∇ × F) · dS = ∮(0i + 0j + 1k) · (k dA)
= ∮dA
= Area of the disk.
The area of the disk x^2 + y^2 < 4 is 4π, so the surface integral is 4π.
Next, let's calculate the line integral:
∮F · dr = ∮(yi + zj + xk) · (dx i + dy j + dz k)
= ∮(y dx + z dy + x dz).
Using the parameterization x = rcosθ, y = rsinθ, and z = 0, with r ranging from 0 to 2 and θ ranging from 0 to 2π, we have:
∮F · dr = ∫∫(0 dx + 0 dy + (rcosθ) dz)
= ∫∫0 dz
= 0.
Therefore, the line integral is 0.
Since the surface integral is 4π and the line integral is 0, we have verified Stokes' theorem for the field F = yi + zj + xk over the disk x^2 + y^2 < 4, z = 0.