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$$
\mbox{In spherical coordinates,}\quad
\delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad
\mbox{such that}
$$
\begin{align}
\color{#66f}{\large\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}}
&=\int_{0^{-}}^{\infty}\dd r\,r^{2}\int_{0}^{\pi}\dd\theta\,\sin\pars{\theta}
\int_{0}^{2\pi}\dd\phi\,{\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}
\\[3mm]&=\underbrace{\bracks{\int_{0^{-}}^{\infty}\delta\pars{r}\,\dd r}}
_{\ds{=\ 1}}\
\underbrace{\bracks{%
\int_{0}^{\pi}\delta\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta}}
_{\ds{=\ 1}}\
\underbrace{\bracks{\int_{0}^{2\pi}\delta\pars{\phi}\,\dd\phi}}_{\ds{=\ 1}}\
\\[3mm]&=\ \color{#66f}{\Large 1}
\end{align}
$$\mbox{Note that}\quad
\int_{{\mathbb R}^{3}}\delta\pars{\vec{r} - \vec{r}_{0}}\,\dd^{3}\vec{r}
=\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}
$$
The "definition" you give of $\delta$ is not mathematically rigorous. The correct definition is that $\langle \delta, \phi \rangle = \phi(0)$ for all $\phi\in C^\infty_c(\mathbb{R})$ or $\phi\in \mathcal{S}(\mathbb{R})$ when working with Fourier transforms. Here $\langle u, \phi \rangle$ is a pairing of a distribution $u$ with a test function $\phi,$ returning a number. It's a bit similar to inner products, but the objects are of different types. You can think of it as the integral $\int_{\mathbb{R}} u(x)\,\phi(x)\,dx.$
An ordinary function $f$ is considered as a distribution by
$\langle f, \phi \rangle := \int f(x) \, \phi(x) \, dx.$
The formula $\delta(\omega)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{i\omega t}\,dt$ is also not rigorous, but I would say that it is at least somewhat better as it can be interpreted within the theory of distributions as the inverse Fourier transform of constant function $\mathbf{1}(t) = 1.$
Fourier transforms of distributions are defined by moving the transform to the test function: $\langle \mathcal{F}u, \varphi \rangle = \langle u, \mathcal{F}\varphi \rangle.$ From this and the definition of $\delta$ above we get
$$
\langle \mathcal{F}\delta, \phi \rangle
= \langle \delta, \mathcal{F}\phi \rangle
= \mathcal{F}\phi(0)
= \left. \int \phi(t) e^{-i\omega t} \, dt \right|_{\omega=0}
= \int \phi(t) \, dt
= \langle \mathbf{1}, \phi \rangle.
$$
Now,
$$
\langle \int_{\mathbb{R}} e^{i\omega t} dt, \phi(\omega) \rangle
= \langle \mathcal{F}\mathbf{1}(-\omega), \phi(\omega) \rangle
= \langle \mathbf{1}(-\omega), \mathcal{F}\phi(\omega) \rangle
= \langle \mathbf{1}(\omega), \mathcal{F}\phi(-\omega) \rangle
\\
= \langle \mathcal{F}\delta(\omega), \mathcal{F}\phi(-\omega) \rangle
= \langle \delta(\omega), \mathcal{F}\mathcal{F}\phi(-\omega) \rangle
= \langle \delta(\omega), 2\pi\phi(\omega) \rangle
= \langle 2\pi\delta(\omega), \phi(\omega) \rangle
,
$$
for every test function $\phi$ so $\int_{\mathbb{R}} e^{i\omega t} dt=2\pi\delta(\omega)$ where the left hand side has been interpreted as $\mathcal{F}\mathbf{1}(-\omega).$
Thus, the integral itself is not rigorous, but there is a rigorous interpretation of it.
Best Answer
One needs to insert a gradient $${\rm Vol}(\{x \in\mathbb{R}^n|f(x)=0\})~=~ \int_{\mathbb{R}^n} \!\mathrm{d}^nx~ \color{red}{|\nabla f(x)|} \delta(f(x)) \tag{1}$$ for purely geometric reasons to ensure that the volume doesn't depend on the parametrization of the constraint function $f$.
Example: The circumference of the unit-circle $\mathbb{S}^1=\{x \in\mathbb{R}^2|f(x)=0\}$. Here the constraint function is $f(x)=x_1^2+x_2^2-1$. Then $\color{red}{|\nabla f(x)|=2}$ for $x\in \mathbb{S}^1$. Hence $${\rm Vol}(\mathbb{S}^1)~\stackrel{(1)}{=}~ \int_0^{2\pi} \!\mathrm{d}\phi\int_0^{\infty} \!r\mathrm{d}r~ \color{red}{2} \delta(r^2-1)~=~2\pi. \tag{2}$$
See also e.g. this related Phys.SE post.