I recently encountered this integral:
$\mathbf J_1 = \int_{0}^{2\pi} I \delta (z) \delta (y-a(1-\cos t))\delta (x-a\sin t) (\mathbf {\hat x} \sin t + \mathbf {\hat y} \cos t) adt$
The way I attempted to solve it was by noting that $\delta (x-a(1-\cos t)) \delta (y-a\sin t)$ is nonzero iff $t = \cos^{-1}(1 - \frac{x}{a}) = \sin^{-1}(\frac{y}{a})$, resulting in the integral evaluating to the following expression
$\mathbf J_1 = Ia\delta (z) (\mathbf {\hat x} \frac{y}{a}+ \mathbf {\hat y} (1 - \frac{x}{a}))$
However, in the context of the problem, that does not make sense. $\mathbf J_1$ is supposed to be the current density of a circular current loop in the xy-plane. So I know that I solved the integral incorrectly. But how? What is $\delta (x-a(1-\cos t)) \delta (x-a\sin t)$ supposed to evaluate to?
Assuming that the third Delta is $\delta(y-a\sin t)$ instead of $\delta(x-a\sin t)$. $J_1$ is already in the simplest form, it is the current density going through the xy-plane $z=0$ in that circle. You can use it to get a current flux through any surface, the t parameter can then help you check all the points on the circle to see if there's an intersection with the surface. If there is one that you take the dot product of the tangent vector to the circle with the surface normal there.