Exercise 1: Worked solutions

1a · Surface coordinates

1. In cylindrical coordinates, ρ measures distance from the axis, φ gives the angle around it, and z gives height. A point has position:

   r = ρ e_ρ + z e_z

   The angle has no term of its own: it sits in the direction of e_ρ = cos φ e_x + sin φ e_y, which turns as you go round the axis. So r still depends on all three coordinates.

   To describe a surface, hold one coordinate fixed and let the other two run over the part of the surface we want.

2. Base: every point is at height z = 0. The disk reaches from the axis to the wall, so:

   z = 0; 0 ≤ ρ ≤ a; 0 ≤ φ < 2π

3. Curved side: every point is at radius ρ = a. Height runs from the base to the top:

   ρ = a; 0 ≤ z ≤ h; 0 ≤ φ < 2π

4. Top: every point is at height z = h. It is another full disk:

   z = h; 0 ≤ ρ ≤ a; 0 ≤ φ < 2π

5. Result: as position vectors, with the ranges above:

   base: r = ρ e_ρ

   side: r = a e_ρ + z e_z

   top: r = ρ e_ρ + h e_z

   In coordinates (ρ, φ, z) these are (ρ, φ, 0), (a, φ, z) and (ρ, φ, h).

   Check: only the side fixes the radius. Each disk includes the centre and reaches all the way to a.

1b · Outward normals

1. A surface normal is perpendicular to the surface. “Outward” chooses between the two perpendicular directions by pointing away from the solid. Cylindrical coordinate unit vectors already have length 1.

2. At the base, the interior lies above the plane z = 0. The outward direction is therefore downward: −e_z.

3. At the top, the interior lies below z = h. The outward direction is upward: +e_z.

4. On the curved side, the solid occupies ρ < a. Increasing ρ moves away from the axis and out of the cylinder, so the outward normal is +e_ρ.

5. Result: the outward unit normals are:

   base −e_z; side +e_ρ; top +e_z

   Check: at either sharp rim, the base/top normal and side normal disagree. The rim has no single smooth-face normal; normals are defined face by face.

2a · Curved surface plus base

1. Use spherical coordinates with polar angle θ measured down from +z. The curved hemisphere lies at r = a, with 0 ≤ θ ≤ π/2 and 0 ≤ φ < 2π.

2. A small surface patch grows with both directions around the sphere. Its area element is:

   dA = a² sin θ dθ dφ

3. Integrate over the curved part:

   A_curve = ∫₀^2π ∫₀^π/2 a² sin θ dθ dφ

4. The angular integral gives 1 and the azimuthal integral gives 2π:

   ∫₀^π/2 sin θ dθ = [−cos θ]₀^π/2 = 1;   A_curve = 2πa²

5. The flat base is a disk of radius a. Add the boundary pieces:

   A_total = 2πa² + πa² = 3πa²

   Check: the curved half is half of 4πa². The area has units of length squared. The shared circular edge has zero area, so it does not affect the sum.

2b · Spherical volume integral

1. Inside the hemisphere, radius runs from the centre to the surface: 0 ≤ r ≤ a. The angles cover 0 ≤ θ ≤ π/2 and 0 ≤ φ < 2π.

2. The spherical-coordinate volume element includes the coordinate Jacobian, which accounts for angular patches widening away from the origin:

   dV = r² sin θ dr dθ dφ

3. Set up the integral. The factors depend on separate variables:

   V = ∫₀^2π ∫₀^π/2 ∫₀^a r² sin θ dr dθ dφ

4. Evaluate each factor:

   ∫₀^a r² dr = a³3;   ∫₀^π/2 sin θ dθ = 1;   ∫₀^2π dφ = 2π

5. Result:

   V = a³3(1)(2π) = 2πa³3

   Check: it is half the full sphere’s 4πa³/3 volume, and its units are length cubed.

3a · Uniform field

1. Parameterize the circle by its angle:

   r(φ) = b cos φ e_x + b sin φ e_y,   0 ≤ φ ≤ 2π

2. Differentiate with respect to the path parameter to get the tangent line element:

   ds = (−b sin φ e_x + b cos φ e_y) dφ = b e_φ dφ

3. Take the dot product with the constant field:

   H · ds = −b H₀ sin φ dφ

4. Integrate over the whole circle:

   ∮ H · ds = −b H₀ ∫₀^2π sin φ dφ = b H₀[cos φ]₀^2π

5. Result:

   ∮ H · ds = 0

   The positive and negative tangential contributions cancel over one full turn.

   Check: the zero is caused by cancellation around the loop, not by the field being perpendicular to the path at every point.

3b · Radial field

1. Use the same circle parameterization. At radius b, the tangent line element is:

   ds = b e_φ dφ

2. The radial unit vector e_ρ points out from the origin, while e_φ follows the circle. In cylindrical coordinates these perpendicular directions satisfy e_ρ · e_φ = 0.

3. The dot product is zero at each point because the directions are perpendicular:

   H · ds = H₀e_ρ · b e_φ dφ = 0

4. Result:

   ∮ H · ds = 0

   Check: unlike 3a, the integrand is zero point by point. No cancellation between different parts of the circle is needed.

3c · Tangential field

1. The path is the circle ρ = b, with φ increasing:

   ds = b e_φ dφ

2. Substitute ρ = b into the field. Its magnitude is constant along this circle:

   H = H₀bae_φ

3. Both vectors point along e_φ, whose dot product with itself is 1:

   H · ds = H₀b²a dφ

4. Integrate the constant around one revolution:

   ∮ H · ds = ∫₀^2π H₀b²a dφ = 2πH₀b²a

5. Result:

   ∮ H · ds = 2πH₀b²a

   It is positive for the assumed counterclockwise direction.

   Check: H₀ has units A/m, and b²/a has units m, so the integral has units A. Reversing the loop makes it negative.