Q:

Find a compact orientable surface whose boundary is the Hopf Link. 

Consider the Hopf link given as 
$$ H_l = S^1 \times \{0\} \quad \cup \{ (x,0,z) \in \mathbb{R}^3 | (x-1)^2 + z^2 = 1\} $$
Find a compact orientable surface $\Sigma$ in $\mathbb{R}^3$ such that $\partial \Sigma = H_l$.

I can construct such a surface in principle using the Seifert algorithm, but I have little idea as to how to explicitly write the surface.  
My idea is to join the left most point of the left circle to the topmost point of the right circle with a line and then as we rotate around the left circle join the corresponding point of the right circle.  However then I have to prove such a set is a surface, is compact and orientable.  Am I on the right track?  How would I go about this? 
Is there a better way?   

A:

If you are familiar with how a Möbius strip is from taking a rectangular strip and gluing two opposite ends together by a 180 degree twist, if you instead do a 360 degree twist you have the Seifert surface of a Hopf link.   Notice that both boundary components are unknots, and the 360 degree twist means the unknots have linking number $\pm 1$ (depending on the way you twisted the strip). 
One perspective on this is through framed links.   Framing is a consistent choice of tangent frame (the tangent, normal, and binormal vectors) along each component of a link, up to a notion of isotopy.   Since the binormal vector can be computed from the tangent and normal vectors, framed links tend to be thought of as made of ribbon (embedded $S^1\times [0,1]$'s) where the $[0,1]$ direction records the normal vector.   A pushoff of a framed link is a boundary component of a ribbon.   
A framing can be assigned a numerical value per link component, the set of which uniquely determines the framing.   Assign to a component the linking number between it and its pushoff. 
An unknot with $\pm 1$ framing (i.e., an unknotted ribbon with a 360 degree twist) gives a Hopf link through the ribbons two boundary components. 
(While we're here, the $0$ framing of a link is from taking a Seifert surface for each component individually then cutting off the boundary ribbon of each surface.  
 Fun fact/exercise: a knot is the unknot if and only if the boundary components of the $0$-framed ribbon form a split link.) 
Another way to get a Seifert surface for the Hopf link is to use the Hopf fibration, which is a smooth map $S^3\to S^2$ where every fiber is an unknot, and every pair of fibers is a Hopf link.   If you take a non-self-intersecting path on $S^2$ and look at the preimage, you get a Seifert surface of the type I have already described.   The Hopf map can be defined by thinking of $S^3$ as being a submanifold of $\mathbb{C}^2$, then $(z,w)\mapsto z/w$ gives a map to the Riemann sphere.   We can parameterize $S^3$ by great circles of the form $(re^{i\theta},se^{i\theta})$ with constants $r,s$ satisfying $|r|^2+|s|^2=1$.  We can assume $s\in\mathbb{R}$ and non-negative by putting it into polar form and combining it with $e^{i\theta}$ as necessary (reparameterizing the $\theta$ in the process), and thus $s=\sqrt{1-|r|^2}$.  Notice that the Hopf map then is $(re^{i\theta},\sqrt{1-|r|^2}e^{i\theta})\mapsto \frac{r}{\sqrt{1-|r|^2}}$.  In real coordinates, this is (with $r=a+bi$ and $a^2+b^2\leq 1$)
$$(a\cos\theta-b\sin\theta,b\cos\theta+a\sin\theta,\sqrt{1-a^2-b^2}\cos\theta,\sqrt{1-a^2-b^2}\sin\theta)\mapsto (\frac{a}{\sqrt{1-a^2-b^2}},\frac{b}{\sqrt{1-a^2-b^2}})$$
Given a particular value of this function $(c,d)$, we can solve for a fiber:
$$\frac{1}{\sqrt{1+c^2+d^2}}(c\cos\theta-d\sin\theta,c\sin\theta+d\cos\theta,\cos\theta,\sin\theta).$$
A stereographic projection of this fiber to $\mathbb{R}^3$ is
$$\frac{1}{\sqrt{1+c^2+d^2}-\sin\theta} ( c\cos\theta-d\sin\theta, c\sin\theta+d\cos\theta, \cos\theta)$$
If we take the stereographically projected fibers for values $(c,d)$ running along a straight-line path from $(2,0)$ to $(4,0)$, then we get what I claimed we would:

An interesting case is when one of the two fibers runs through $(0,0,0,1)$, thereby running through infinity in the stereographic projection.   These pictures come from the fibers of a path from $(0,0)$ to $(4,0)$:

One can imagine this as the limit of taking one of the unknots and making its radius get arbitarily large, pulling the Seifert surface along with it. ( Note that the surface is being cut off by the bounding box --- it meets infinity sort of along a semicircle.) 
Here are some views of the surface for the path $(c,d)=(\cos\alpha,\sin\alpha)$ for $\alpha\in[0,\pi]$:

The full loop around the origin gives a torus (the Clifford torus).   This particular example shows one of the two types of Villarceau circles on a standard torus in $\mathbb{R}^3$.
For fun, here are portions of the surfaces for $(c,d)=(1.5^r\cos\alpha,1.5^r,\sin\alpha$ for $\alpha\in[0,\pi]$ and $r=\{-2,1,0,1,2\}$:

