Flux of the second kind over the paraboloid x²+y²+2z = 4

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

Problem

Evaluate the surface integral of the second kind:

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

where S is the upper side of the part of the paraboloid x² + y² + 2z = 4, z ≥ 0.

Solution

Step 1

Denote the components of the vector field:

P = x^{2}, Q = y^{2}, R = 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 paraboloid x² + y² + 2z = 4 intersects the Oxy plane at z = 0 along the circle x² + y² = 4 of radius 2. Close the surface S by the disk D in the plane z = 0 (x² + y² ≤ 4) with the outward normal directed downward. Then Σ = S ∪ D is a closed surface.

Why close with a disk in the plane z = 0?

The boundary of the surface S is the circle x² + y² = 4 in the plane z = 0. The natural closure is the flat disk spanning this circle. Together S and D form a closed surface bounding the body V under the paraboloid.

Step 4

Apply the Ostrogradsky-Gauss theorem to the closed surface Σ:

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

Step 5

By symmetry of the body V with respect to the planes x = 0 and y = 0:

Why is ∭ x dV = ∭ y dV = 0?

The region V is symmetric with respect to the plane x = 0: each point (x, y, z) has a counterpart (−x, y, z) with the same density. Their contributions to the integral of x cancel out. Similarly for y. The function z does not have such symmetry, hence its integral is non-zero.

\underset{V}{∭} x d V = \underset{V}{∭} y d V = 0

Step 6

It remains to compute ∭_V z dV. Change to cylindrical coordinates:

\underset{V}{∭} z d V = \int_{0}^{2 π} d φ \int_{0}^{2} r d r \int_{0}^{(4 - r^{2}) / 2} z d z

\int_{0}^{(4 - r^{2}) / 2} z d z = \frac{{(4 - r^{2})}^{2}}{8}

Step 7

Evaluate the inner integral over z, then the middle integral over r with the substitution u = 4 − r²:

\int_{0}^{2} r \cdot \frac{{(4 - r^{2})}^{2}}{8} d r = \frac{4}{3}

How is the substitution u = 4 − r² performed?

With u = 4 − r² we have du = −2r dr, hence r dr = −du/2. The limits: for r = 0, u = 4; for r = 2, u = 0. The swap of limits together with the minus sign turns the integral into ∫₀⁴ u² du / 2.

Step 8

The outer integral over φ gives the factor 2π:

\underset{V}{∭} z d V = \frac{4}{3} \cdot 2 π = \frac{8 π}{3}

\underset{V}{∭} (2 x + 2 y + 2 z) d V = 2 \cdot \frac{8 π}{3} = \frac{16 π}{3}

Step 9

Evaluate the integral over the disk D. On D: z = 0, hence R = z² = 0 and the whole integral vanishes:

\iint_{D} = - \iint_{D} z^{2} d x d y = 0

Why is the integral over D equal to zero?

On the disk D the coordinate z is identically zero, so R = z² = 0. The outward normal is n = (0, 0, −1), hence the scalar product (F, n) = −R = 0. Therefore the disk D contributes nothing to the flux.

Step 10

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

\iint_{S} = \frac{16 π}{3} - 0 = \frac{16 π}{3}

Answer

16π/3.

Frequently asked questions

How do I evaluate the flux through the upper side of a paraboloid?

The paraboloid surface is not closed, so close it with a disk in the plane z = 0. Apply the Ostrogradsky-Gauss theorem to the closed surface, evaluate the triple integral of the divergence, and then subtract the integral over the closing disk.

Why is the answer here larger than in problem 9.3 (3π/4)?

In problem 9.3 the surface is a hemisphere of radius 1 with volume V = 2π/3. Here the paraboloid x² + y² + 2z = 4 bounds a larger body (base radius 2, height 2). Its volume and the integral of z are correspondingly larger, giving the larger answer 16π/3.

How do I choose the order of integration in cylindrical coordinates?

Usually one integrates in the order: first over z (from the lower to the upper boundary), then over r (from 0 to the boundary of the projection), then over φ (from 0 to 2π). This order is convenient when the body is bounded above by a surface z = f(x, y) and below by the plane z = 0.

What if the divergence of the field is complex?

If the divergence contains terms antisymmetric with respect to coordinate planes, their contributions vanish due to the symmetry of the region. This significantly simplifies the computation. In our problem this is how ∭ x dV and ∭ y dV vanish.