Flux of the second kind over a hemisphere of radius 1

This problem asks to evaluate a surface integral of the second kind over the upper side of a hemisphere, using the Ostrogradsky-Gauss theorem.

Problem

Evaluate the surface integral of the second kind:

\iint_{S} (x^{2} + y^{2}) d y d z + (x^{2} + z^{2}) d z d x + (y^{2} + z^{2}) d x d y

where S is the upper side of the hemisphere x² + y² + z² = 1, z ≥ 0.

Solution

Step 1

Denote the components of the vector field:

P = x^{2} + y^{2}, Q = x^{2} + z^{2}, R = y^{2} + z^{2}

Step 2

Compute the divergence of the vector field:

div \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 2 x + 2 y + 2 z

Step 3

The surface S is not closed. Close it by the disk D in the plane z = 0 (x² + y² ≤ 1). The outward normal to the hemisphere on D is directed downward: n = (0, 0, −1). Then Σ = S ∪ D is a closed surface.

Why choose the disk in the plane z = 0?

The hemisphere x² + y² + z² = 1, z ≥ 0 has a boundary — the circle x² + y² = 1 in the plane z = 0. The natural closure is the flat disk in the same plane. Together they form a closed surface bounding a semi-ball volume V.

Step 4

Apply the Ostrogradsky-Gauss theorem to the closed surface Σ = S ∪ D:

\iint_{S} + \iint_{D} = \underset{V}{∭} (2 x + 2 y + 2 z) d V

Step 5

Evaluate the triple integral. By symmetry of the region V with respect to the planes x = 0 and y = 0:

Why is ∭ z dV = π/4?

The volume of a hemisphere of radius 1 is (2/3)π. The centroid of a hemisphere along the z axis is at height 3/8 from the base (a standard result). Hence ∭_V z dV = V · z_c = (2π/3)·(3/8) = π/4.

\underset{V}{∭} x d V = \underset{V}{∭} y d V = 0, \underset{V}{∭} z d V = \frac{π}{4}

\underset{V}{∭} (2 x + 2 y + 2 z) d V = 2 \cdot \frac{π}{4} = \frac{π}{2}

Step 6

Evaluate the integral over the disk D. On D: z = 0, so R = y² + z² = y². The outward normal points downward (n = (0,0,−1)), giving a minus sign:

\iint_{D} = - \iint_{D} y^{2} d x d y = - \int_{0}^{2 π} {sin}^{2} φ d φ \int_{0}^{1} r^{3} d r = - π \cdot \frac{1}{4} = - \frac{π}{4}

Why does a minus sign appear before the integral over D?

In the formula ∬_D P dy dz + Q dz dx + R dx dy = ∬_D (F, n) dσ the normal n is chosen outward to the region V. On the lower face (the plane z = 0) the outward normal points downward, i.e. n = (0, 0, −1). Hence the scalar product (F, n) = −R = −y², which gives the minus sign.

Step 7

By additivity: ∬_S = ∭_V − ∬_D. Substitute the found values:

\iint_{S} = \frac{π}{2} - (- \frac{π}{4}) = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4}

Answer

3π/4.

Frequently asked questions

How do I evaluate the flux of a vector field through a non-closed surface?

If the surface is not closed, close it with an additional surface (usually flat), obtaining a closed surface. Then apply the Ostrogradsky-Gauss theorem to the closed surface, evaluate the integral over the additional part separately, and subtract it.

What is the Ostrogradsky-Gauss theorem?

The theorem relates the flux of a vector field through a closed surface to the triple integral of the field divergence over the volume bounded by the surface: ∬_Σ (F, n) dσ = ∭_V div F dV. Here n is the outward normal to the surface.

How do I choose the direction of the normal?

For a closed surface bounding a region V, the normal is always taken outward (directed out of V). On the upper part (hemisphere) — upward, on the lower part (disk) — downward. This corresponds to the direction "out of the body".

Why is the integral over the disk D not zero?

On the disk D: z = 0, so R = y² + z² = y². The integral ∬_D y² dx dy over the unit disk equals π/4 — not zero, because the integrand y² is odd only in y, not in x. Hence the disk contributes significantly to the flux.