I am trying to evaluate the following integral (3-d delta function):
$$\int (10x^2 +3y^3-4z^2+8)\delta(2\vec{r})d^3\vec{r}$$
Not sure where to proceed, I know that $$\delta^3(\vec{r}) = \delta(x)\delta(y)\delta(z)$$
$$\delta(\alpha x) = \frac{1}{|\alpha|^n}\delta(x)$$
$$\begin{equation} \delta^3(\vec{r}-\vec{r}_0)=\delta(x-x_0)\,\delta(y-y_0)\,\delta(z-z_0) \end{equation} \implies \int\delta(2x)dx\int\delta(2y)dy\int\delta(2z)dz $$ $$= \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
but then substituting in the function $(10x^2 +3y^3-4z^2+8)$ will the answer then be $1$?
Yes it is $1$ because of another delta property:
$$\delta(\alpha x) = \frac{1}{|\alpha|^n}\delta(x)$$
for a $\delta$ in $\mathbb{R}^n$