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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit/8/c/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 8 / c / Solution by

Codex  0 2026-10-25

Use the outward normal on the spherical patch. It is the quarter of the spherical surface lying below both the horizontal plane and the plane

y = 0

, with both signs of

x

included. Close it by the two half-discs

D_{y}

in

y = 0, z \leq 0

and

D_{z}

in

z = 0, y \leq 0

.

Figure 1.
The spherical patch of radius four with y and z nonpositive, shown with the two planar caps used to close the volume.
The colored curved part is the requested surface. The planar half-discs are auxiliary surfaces for the divergence theorem; their outward normals point in positive y and positive z. Details.

The divergence of the polynomial vector field is

\nabla \cdot F = 4 x y

. Its volume integral over the enclosed quarter-ball vanishes by oddness in

x

. Thus the divergence theorem gives

Φ_{S} + Φ_{D_{y}} + Φ_{D_{z}} = 0

.

On

D_{y}

, the outward normal is

+ \overset{y}{^}

and

F_{y} = - 3 z^{4}

. On

D_{z}

, it is

+ \overset{z}{^}

and

F_{z} = x^{3} + y^{2}

; the odd term integrates to zero. The needed half-disc moments for radius

R

are

\int_{D_{y}} z^{4} d x d z = (\int_{0}^{R} r^{5} d r) (\int_{π}^{2 π} sin^{4} ϕ d ϕ) = \frac{π R ^{6}}{16},

and

\int_{D_{z}} y^{2} d x d y = (\int_{0}^{R} r^{3} d r) (\int_{π}^{2 π} sin^{2} ϕ d ϕ) = \frac{π R ^{4}}{8} .

At

R = 4

, the cap fluxes are

- 768 π

and

32 π

, respectively. Therefore the flux through a spherical quadrant by planar caps is

Φ_{S} = 768 π - 32 π = 736 π .

The source does not explicitly specify a normal for this part; the boxed answer adopts the usual outward spherical orientation. Choosing the inward normal reverses its sign.

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