Example: Dirac function in two dimensions. I am considering an equation of the form $$\int_{\mathbb{R}^{2}} f(x) \delta(s - x \cdot \theta)dx$$
where $\delta$ is a two-dimensional Dirac function. What does this evaluate to, exactly? I know that if we instead had
$$\int_{\mathbb{R}^{2}} f(x) \delta(s - x)dx$$
then this would evaluate to $f(s)$. But what effect does the dot product have on this evaluation?
 A: First, let's clear up some potential meaning confusion.  In the equation
$$\int_{\Bbb{R}^2} g(\vec{x}) \delta(\vec{y}-\vec{x}) dx = g(\vec{y}) $$
the nature of $g(x)$ is $g : \Bbb{R}^2 \mapsto \Bbb{R}$.
 But in the expression
$$\int_{\Bbb{R}^2} f(\vec{x}) \delta(\vec{s}-\vec{x} \cdot \vec{\theta}) dx  $$
the nature of $f(x)$ is $f: \Bbb{R} \mapsto \Bbb{R}$,
that is, $f(x)$ is a mapping taking a scalar into a scalar, not a vector into a scalar.
Now to get the answer.  Say we rotate our coordinate variables, to a frame where $\vec{\theta} = p \hat{x}'$. The Hessian of this transformation is one, and in the new coordinates 
$\vec{x} \cdot \vec{\theta} = p x'$.
So the integral becomes
$$
\int_{y' = -\infty}^{\infty} \int_{x' = -\infty}^{\infty} f(x',y') \delta(s-px') dx \, dy 
= \int_{y' = -\infty}^{\infty} \frac{1}{p} f(\frac{s}{p},y') dy
$$
Now we want to go back to the original frame, but again since the rotation did not change any scales, one unit of distance along $y'$ is the same as one unit of distance along the line $\vec{x}\cdot\vec{\theta} = s$.  So if we let $\hat{u}$ be a unit vector parallel to $\theta$, and take $\vec{x}_0$ to be any point along  $\vec{x}_0 \cdot \vec{\theta} = s$, we will be able to express the line integral as an integral along points of the form $\vec{x}_0 + t\hat{u}$ with unit distance scaling. 
So, using the fact that $p = |\vec{\theta}|$, the answer is the line integral
$$
\frac{1}{|\vec{\theta}|}\int_{t=-\infty}^\infty f(\vec{x}_0 + t\hat{u})\, dt
$$
The important non-trivial thing to note is that the weighting along the line does not depend on the value or the derivatives of the function $f$.
