= Solution
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 $y=0,z\leq0$ and $D_z$ in $z=0,y\leq0$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit-spherical-quadrant.png]
{title=The spherical patch of radius four with y and z nonpositive, shown with the two planar caps used to close the volume.}
{description=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.}
The <divergence> of the polynomial <vector field> is $\nabla\cdot F=4xy$. Its <volume integral> over the enclosed quarter-ball vanishes by oddness in $x$. Thus the <divergence theorem> gives $\Phi_S+\Phi_{D_y}+\Phi_{D_z}=0$.
On $D_y$, the outward normal is $+\hat y$ and $F_y=-3z^4$. On $D_z$, it is $+\hat z$ and $F_z=x^3+y^2$; the odd term integrates to zero. The needed half-disc moments for radius $R$ are
$$
\int_{D_y}z^4\,dx\,dz
=\left(\int_0^Rr^5\,dr\right)\left(\int_\pi^{2\pi}\sin^4\phi\,d\phi\right)
=\frac{\pi R^6}{16},
$$
and
$$
\int_{D_z}y^2\,dx\,dy
=\left(\int_0^Rr^3\,dr\right)\left(\int_\pi^{2\pi}\sin^2\phi\,d\phi\right)
=\frac{\pi R^4}{8}.
$$
At $R=4$, the cap <fluxes> are $-768\pi$ and $32\pi$, respectively. Therefore the <flux through a spherical quadrant by planar caps> is
$$
\boxed{\Phi_S=768\pi-32\pi=736\pi.}
$$
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.
Back to article page
