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, we need to compute the dot product of the vector field F and the oriented normal vector dS. Here's how we can do it step by step:
1. Determine the parametrization of the surface S:
In this case, S is the part of the cone z = sqrt(x^2 + y^2) beneath the plane z = 3. To parametrize this surface, we can use cylindrical coordinates:
x = r cosθ
y = r sinθ
z = r
2. Compute the partial derivatives of the parametrization:
∂r/∂x = cosθ
∂r/∂y = sinθ
∂r/∂z = 1
∂θ/∂x = -r sinθ
∂θ/∂y = r cosθ
∂θ/∂z = 0
3. Calculate the cross product of the partial derivatives:
∂r/∂x × ∂θ/∂x = (cosθ) × (-r sinθ) = -r cos^2θ
∂r/∂y × ∂θ/∂y = (sinθ) × (r cosθ) = r sin^2θ
∂r/∂z × ∂θ/∂z = (1) × (0) = 0
4. Find the magnitude of the cross product:
|∂r/∂x × ∂θ/∂x, ∂r/∂y × ∂θ/∂y, ∂r/∂z × ∂θ/∂z| = |-r cos^2θ, r sin^2θ, 0| = r(cos^2θ + sin^2θ) = r.
5. Normalize the cross product vector to get the outward unit normal vector dS:
dS = (1/|∂r/∂x × ∂θ/∂x, ∂r/∂y × ∂θ/∂y, ∂r/∂z × ∂θ/∂z|) * (∂r/∂x × ∂θ/∂x, ∂r/∂y × ∂θ/∂y, ∂r/∂z × ∂θ/∂z) = (1/r)(-r cos^2θ, r sin^2θ, 0) = (-cos^2θ, sin^2θ, 0).
6. Calculate the dot product between F and dS:
F · dS = (x i + y j + z^4 k) · (-cos^2θ i + sin^2θ j + 0 k) = -x cos^2θ + y sin^2θ + z^4 * 0 = -x cos^2θ + y sin^2θ.
7. Substitute the parametrization values into the dot product expression:
F · dS = -(r cosθ) cos^2θ + (r sinθ) sin^2θ = -r cos^3θ + r sin^3θ.
8. Set up the integral for the flux:
∬S F · dS = ∫∫R (-r cos^3θ + r sin^3θ) r dr dθ,
Where R is the region in the rθ-plane that corresponds to the surface S.
9. Evaluate the surface integral:
We need to determine the bounds of integration for r and θ based on the given surface. In this case, the cone is beneath the plane z = 3, so the height of the cone is z = 3 - r, and the bounds for r will be from 0 to 3. The bounds for θ will be from 0 to 2π, as we want to cover the entire surface.
∬S F · dS = ∫0^3 ∫0^2π (-r cos^3θ + r sin^3θ) r dθ dr.
You can evaluate this double integral numerically to find the flux of F across the given surface S.