Six surface flux questions with explained solutions

Parameterize by $\mathbf r(x,y)=(x,y,1+x+2y)$. The z component of $\mathbf r_x\times\mathbf r_y$ should be positive. The parameter rectangle has area 2, but the tilted patch has a different area.

The tangent vectors are $\mathbf r_x=(1,0,1)$ and $\mathbf r_y=(0,1,2)$. Their cross product is $(-1,-2,1)$, which points upward. The signed integrand is

Thus

The unit normal is $(-1,-2,1)/\sqrt6$, and the surface area is $2\sqrt6$. Since the field and normal are constant, the second calculation gives $(3/\sqrt6)(2\sqrt6)=6$. Keeping the unit normal but using only the projected area 2 would omit the area stretch. Downward orientation gives $-6$.

Before moving on: identify the exact factor lost by the incorrect unit-normal calculation. Repair it with the area-vector explanation.

Use $(-2x,-2y,1)\,dx\,dy$ for the upward vector area element. Change the projection-disk integral to polar coordinates after taking the dot product.

Substitution into the field and the dot product give

With $x=\rho\cos\theta$, $y=\rho\sin\theta$, the planar Jacobian is $\rho$. Therefore

The upward normal contribution is positive for $\rho<1/\sqrt2$ and negative for $\rho>1/\sqrt2$. The inner disk contributes $\pi/4$; the surrounding annulus contributes $-\pi/4$. They cancel. The field is never zero because its third component is 1, and it is tangent only along the circle where the normal contribution vanishes.

This surface has an edge at $z=1$; it does not enclose a solid by itself. The direct calculation is already valid, so no closing surface is needed for this method. If the Jacobian or area element was unclear, revisit the graph formula before retrying.

Use $\mathbf r(\theta,z)=(2\cos\theta,2\sin\theta,z)$. Check the direction of $\mathbf r_\theta\times\mathbf r_z$. The top and bottom disks have different field values and opposite outward normals.

With $0\le\theta\le2\pi$ and $0\le z\le3$,

so $\mathbf r_\theta\times\mathbf r_z=(2\cos\theta,2\sin\theta,0)$. It points outward and has magnitude 2. The field evaluated on the side is $(2\cos\theta,2\sin\theta,z+1)$, hence the dot product is 4. The vertical component contributes nothing through this side.

For a theorem check, the solid cylinder has volume $\pi(2)^2(3)=12\pi$. The polynomial field has divergence 3, so its whole closed-boundary flux is $36\pi$.

• Top disk: $z=3$, $\mathbf n=\mathbf k$, $\mathbf F\cdot\mathbf n=4$, area $4\pi$, flux $16\pi$.
• Bottom disk: $z=0$, $\mathbf n=-\mathbf k$, $\mathbf F\cdot\mathbf n=-1$, area $4\pi$, flux $-4\pi$.
• Curved side: $36\pi-16\pi-(-4\pi)=24\pi$.

The sum is $24\pi+16\pi-4\pi=36\pi$. Forgetting either disk means that the theorem has been applied to a different boundary. Review signed closing-surface subtraction if those signs did not match your sketch.

At height $z$, the solid's cross-section has area $\pi(4-z^2)$. The closing disk has radius $\sqrt3$, not 2. For the direct calculation, use spherical colatitude $0\le\varphi\le\pi/3$.

The cap and disk enclose the solid inside the sphere with $1\le z\le2$. Its volume is

The field is polynomial and has divergence 3, so the flux through the whole boundary is $5\pi$. On the added disk, the solid is above the plane, hence the outward normal is $-\mathbf k$. At $z=1$, the field's third component is 2. The disk area is $3\pi$, giving

For a direct check, parameterize the sphere by

The cap condition is $2\cos\varphi\ge1$, which explains the upper colatitude bound. The outward unit normal is $(x,y,z)/2$, so

The surface-area element is $4\sin\varphi\,d\varphi\,d\theta$. Therefore

Both methods agree for the curved cap alone. If your direct answer disagrees, check both the sphere radius in the area element and the colatitude limit. Revisit the hemisphere's direct calculation.

The student's volume calculation concerns the boundary of the solid under the paraboloid and above $z=0$. List every surface on that boundary before deciding what $24\pi$ represents.

The volume is correctly calculated:

The field is smooth everywhere, so $24\pi$ is the total outward flux through the paraboloid and its base disk together. The missing step is removing the base contribution. The disk has radius 2, normal $-\mathbf k$, and $\mathbf F\cdot\mathbf n=-1$, giving $\Phi_B=-4\pi$. The upward orientation on the paraboloid is outward from this solid. Thus

For the direct calculation, $g_x=-2x$, $g_y=-2y$, so the upward area-vector density is $(2x,2y,1)$. On the graph, $\mathbf F=(x,y,5-x^2-y^2)$, and the dot product becomes

Therefore

Downward orientation reverses the surface flux to $-28\pi$. The enclosed volume stays the same. The divergence theorem's outward-orientation convention must be applied before that sign reversal. Review the theorem conditions and the closing-surface identity if you treated a correct volume as a complete solution.

On a radius-$R$ sphere, write $\mathbf n=(x,y,z)/R$. For the shell, the normal on its inner boundary points toward the origin, away from the material of the shell.

The field is undefined at the origin. It is smooth on the sphere, but not on an open set containing the filled ball. Therefore the usual divergence theorem cannot be applied to that ball. Removing one point from the integral without accounting for the new boundary is not a justification.

For completeness, write $r=\sqrt{x^2+y^2+z^2}>0$. The three diagonal derivatives sum to

where the field is defined. This calculation is correct; its use over the filled ball was the problem.

On a radius-$R$ sphere,

Multiplying by area $4\pi R^2$ gives outward flux $4\pi$, so the required radius-2 answer is $4\pi$.

The shell excludes the origin and has a smooth field on a neighbourhood of its closure. Its boundary has two components. The outer sphere contributes $4\pi$; the inner sphere uses the opposite radial normal and contributes $-4\pi$. The total $4\pi-4\pi=0$ agrees with the volume integral of zero divergence over the shell. The theorem succeeds on the region that satisfies its hypotheses.

If the sign of the inner boundary surprised you, reread outward orientation and cavities. For the domain failure, revisit the differentiability condition.