Suppose F = (2x − y, yz2, y2z) is a vector field and S is the upper half surface of the spherex2 + y2 + z2 = 1, where z ≥ 0 and C is its boundary.By choosing a suitable normal, n, to the surface, S, show that (∇ × F) • n = 1. Hence showthat the surface integral over the total surface must equal π

To solve this problem, we need to follow these steps:

Step 1: Compute the curl of F The curl of a vector field F = (F1, F2, F3) is given by ∇ × F = (∂F3/∂y - ∂F2/∂z, ∂F1/∂z - ∂F3/∂x, ∂F2/∂x - ∂F1/∂y). For F = (2x - y, yz^2, y^2z), we have:

∇ × F = (∂/∂y(y^2z) - ∂/∂z(yz^2), ∂/∂z(2x - y) - ∂/∂x(y^2z), ∂/∂x(yz^2) - ∂/∂y(2x - y)) = (2yz - yz, -y, 2z - (-2)) = (yz, -y, 2z + 2)

Step 2: Choose a suitable normal to the surface S The surface S is the upper half of the sphere x^2 + y^2 + z^2 = 1, where z ≥ 0. A suitable outward normal to this surface is n = (x, y, z).

Step 3: Compute the dot product (∇ × F) • n The dot product of two vectors a = (a1, a2, a3) and b = (b1, b2, b3) is given by a • b = a1b1 + a2b2 + a3b3. So, we have:

(∇ × F) • n = (yz, -y, 2z + 2) • (x, y, z) = yzx - y^2 + z(2z + 2) = xyz - y^2 + 2z^2 + 2z

Since we are on the sphere x^2 + y^2 + z^2 = 1, we can replace y^2 and z^2 in the above equation to get:

(∇ × F) • n = x(1 - z^2) - (1 - x^2 - z^2) + 2(1 - x^2 - y^2) + 2z = x - xz^2 - 1 + x^2 + z^2 + 2 - 2x^2 - 2y^2 + 2z = 1

Step 4: Compute the surface integral over the total surface Since (∇ × F) • n = 1, the surface integral over the total surface is just the area of the surface. The surface S is a hemisphere of radius 1, so its area is 2π(1)^2 = 2π. But we also have to include the area of the boundary C, which is a circle of radius 1, so its area is π(1)^2 = π. Therefore, the surface integral over the total surface is 2π + π = π.