Evaluate the surface integral
S
F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = x i + y j + z4 k
S is the part of the cone
z = sqrt(x2 + y2) beneath the plane z = 3
with upward orientation
To evaluate the surface integral, or the flux of vector field F across the oriented surface S, we first need to parameterize the surface S.
The given surface is a part of the cone where z = √(x^2 + y^2) is beneath the plane z = 3. We can rewrite this equation as z = 3.
Let's parameterize this surface S using a surface parameterization. We can use cylindrical coordinates (ρ, φ, z), where ρ is the distance from the z-axis, φ is the angle in the xy-plane, and z is the height.
Since z = 3, we can rewrite this as z(ρ, φ) = 3.
Next, we need to determine the bounds for the parameters ρ and φ. The cone opens upwards, and we want to evaluate the flux of F across the part of the cone below the plane z = 3.
For ρ, we can choose the bounds between 0 and 3, since the cone is defined for all ρ ≥ 0.
For φ, we can choose the bounds between 0 and 2π, since we want to cover a full revolution around the z-axis.
Let's now calculate the outward unit normal vector n̂ at each point on the surface. Since the surface is a part of the cone without any holes, the outward unit normal vector points away from the z-axis and in the positive z-direction. Therefore, n̂ = (0, 0, 1).
Now, let's calculate the dot product F · dS, where dS is the differential surface area element vector.
dS = (∂r/∂ρ × ∂r/∂φ) dρ dφ,
where r(ρ, φ) = (ρcosφ, ρsinφ, 3) is the position vector on the surface S.
Taking partial derivatives, we get:
∂r/∂ρ = (cosφ, sinφ, 0),
∂r/∂φ = (-ρsinφ, ρcosφ, 0).
Taking the cross product, we have:
∂r/∂ρ × ∂r/∂φ = (0, 0, ρ).
So, dS = (0, 0, ρ) dρ dφ.
Now, let's evaluate the dot product F · dS:
F · dS = (x i + y j + z^4 k) · (0 i + 0 j + ρ k) dρ dφ
= ρ z^4 dρ dφ.
Substituting z = 3, we have:
F · dS = 3^4 ρ dρ dφ
= 81 ρ dρ dφ.
Finally, let's set up the integral to compute the flux:
∬S F · dS = ∫∫S 81 ρ dρ dφ.
The bounds for the double integral are: 0 ≤ ρ ≤ 3 and 0 ≤ φ ≤ 2π.
Evaluating this double integral will yield the flux of vector field F across the oriented surface S.