Let $D$ denote the unit disk and let $C$ denote the unit circle - i.e. the boundary of $D$. Suppose we hold the temperature of the top half of $C$ to $1$ and the bottom half of $C$ to $0$ and allow the heat to diffuse through $D$. Use the Poisson integral formula
$$u(r,\theta )=\frac{1}{2\pi }\int _{-\pi }^{\pi }f(\varphi )\frac{R^2-r^2}{R^2+r^2-2R r \cos (\theta -\varphi )}d\phi$$
to express the resulting steady state temperature distribution.
Given the information stated in the problem, I can simplify the expression for the steady state temperature distribution. First, $R=1$ because $C$ is the unit circle with radius 1. Also,
$$ f(\phi)=\begin{cases}
0 & -\pi \lt \phi \lt 0 \\
1 & 0 \lt \phi \lt \pi
\end{cases}
$$
because the temperature of the bottom half of the circle $C$ is 0 and the temperature of the top half is 1.
Using the Poisson integral formula (swapping "$\varphi$" for "$\phi$")
$$u(r,\theta)=\frac{1}{2\pi} \int_{-\pi}^{\pi} f(\phi) \frac{R^2-r^2}{R^2+r^2-2Rr\cos(\theta-\phi)}\ d\phi$$
and plugging in for $f(\phi)$ and $R$ gives
$$u(r,\theta)=\frac{1}{2\pi} \int_{-\pi}^{0} (0) \frac{1^2-r^2}{1^2+r^2-2r\cos(\theta-\phi)}\ d\phi$$
$$+\frac{1}{2\pi} \int_{0}^{\pi} (1) \frac{1^2-r^2}{1^2+r^2-2r\cos(\theta-\phi)}\ d\phi$$
or
$$u(r,\theta)=\frac{1}{2\pi} \int_{0}^{\pi} \frac{1-r^2}{1+r^2-2r\cos(\theta-\phi)}\ d\phi$$
This integral could be simplified further, but Dr. McClure advised me to stop at this point on these types of problems (like problem #2 on the Exam II review handout).