Let $X:=\mathbb P(a\_0,a\_1, \ldots, a\_n)$ be a **well formed** weighted projective variety. Let $-K\_X$ be its anticanonical divisor, then how to express its volume ${\rm vol}(-K\_X)=(-K\_X)^n$ in terms of $a\_0,\ldots, a\_n$?


In principle, this could be computed by toric geometry, but the data seems too complicated to compute (especially to write the dual lattice). Besides, I was wondering if there is a generally formula for the self intersection $D^n$ of the torus invariant divisor 
$D = \sum\_{i=0}^n c\_i D\_i$? 



Volume of $-K_X$ for a weighted projective variety
I am just rewriting my comment above as an answer.


Let $S=k[x\_0,\dots,x\_n]$ be the $\mathbb{Z}\_{\geq 0}$-graded $k$-algebra with every $x\_i$ homogeneous of degree $a\_i.$ Denote by $X$ the associated projective $k$-scheme, $X =\text{Proj}\ S.$ Denote by $a$ the least common multiple of $(a\_0,\dots,a\_n).$ For every $i,$ denote by $b\_i$ the integer such that $a\_i\cdot b\_i$ equals $a.$ Denote by $b$ the product $b\_0\dots b\_n.$ 


Let $R=k[y\_0,\dots,y\_n]$ be the $\mathbb{Z}\_{\geq 0}$-graded $k$-algebra with every $y\_i$ homogeneous of degree $a.$ Denote by $Y$ the associated projective $k$-scheme, $Y=\text{Proj}\ R.$ This is $k$-isomorphic to $\mathbb{P}^n\_k.$ There is a unique homomorphism of $\mathbb{Z}\_{\geq 0}$-graded $k$-algebras, $$f^\*:R \to S, \ \ y\_i\mapsto x\_i^{b\_i}.$$ The inverse image of the irrelevant (prime) ideal is primary for the irrelevant (prime) ideal. Thus, there is an induced morphism of $k$-schemes, $$f:X\to Y.$$ This morphism is finite, and it is flat over an open subscheme that includes the open $Y\_\*=D\_+(y\_0\dots y\_n).$ In fact, because of the well-formedness hypothesis, the restriction of the morphism over $Y\_\*$ is naturally a torsor for the finite, flat, commutative group scheme $\Gamma=\mu\_{b\_0}\times \dots \times \mu\_{b\_n}$ acting by $$(\zeta\_0,\dots,\zeta\_n)\cdot [x\_0,\dots,x\_n] = [\zeta\_0\cdot x\_0,\dots, \zeta\_n\cdot x\_n].$$


Assume that $(a\_0,\dots,a\_n)$ is well-formed, i.e., the greatest common divisor of any $n$ of the $n+1$ weights equals $1$. In particular, this implies that the smooth locus $X^o$ of $X$ is a dense open subscheme whose complement has codimension $\geq 2$. Moreover, the Picard group of $X^o$ is generated by the ample invertible sheaf $\mathcal{O}\_X(1)|\_{X^o}$ that is the restriction of the rank $1$, reflexive, coherent sheaf $\mathcal{O}\_X(1) = \widetilde{S[1]}.$ In particular, for every integer $d\geq 0,$ $$H^0(X^o,\mathcal{O}\_X(d)|\_{X^o}) = H^0(X,\mathcal{O}\_X(d)) = S\_d.$$


The Picard group of $Y$ is generated by an ample invertible sheaf $\mathcal{O}\_Y(1)$ whose vector space of global sections is the free $k$-vector space with basis $y\_0,\dots,y\_n.$ Since $f^\*(y\_i)$ has degree $a,$ the pullback $f^\*\mathcal{O}\_Y(1)$ is an ample invertible sheaf on $X$ whose restriction to $X^o$ equals $\mathcal{O}\_X(a)|\_{X^o}.$ In particular, $f^\*\mathcal{O}\_Y(-(a\_0+\dots+a\_n))|\_{X^o}$ is isomorphic to $\omega\_{X^o/k}^{\otimes a}.$ Thus, the $n$-fold self-intersection on $X$ of $c\_1(f^\*\mathcal{O}\_Y(a\_0+\dots +a\_n))$ equals $(a\_0+\dots+a\_n)^n$ times the $n$-fold self-intersection on $X$ of $f^\*c\_1(\mathcal{O}\_Y(1)).$


The $n$-fold self-intersection on $Y$ of $c\_1(\mathcal{O}\_Y(1))$ is the unique class whose cap product with $[Y]$ equals the class of every $k$-point of $Y$. Thus, the $n$-fold self-intersection on $X$ of $f^\*c\_1(\mathcal{O}\_Y(1))$ equals the class of every fiber of $f$ over any element of $Y\_\*.$ Since the morphism is a torsor for the finite, flat, commutative group scheme $\Gamma$ of length $b = b\_0\cdots b\_n,$ it follows that the $n$-fold self-intersection $X$ of $f^\*c\_1(\mathcal{O}\_Y(1))$ equals $b.$


Putting the pieces together, there is a unique invertible $\mathcal{O}\_X$-module, $f^\*\mathcal{O}\_Y(a\_0+\dots + a\_n)$, whose restriction to $X^o$ equals $(\omega\_{X^o/k}^\vee)^{\otimes a}.$ The $n$-fold self-intersection of $c\_1(f^\*\mathcal{O}\_Y(a\_0+\dots+a\_n))$ equals $(a\_0+\dots+a\_n)^n b.$ Thus, considered as a rational number, the $n$-fold self-intersection of $c\_1(\omega\_{X/k}^\vee)$ equals $(a\_0+\dots+a\_n)^nb/a^n.$ Finally, using the fact that $b\_i/a$ equals $1/a\_i,$ this gives, $$\left( c\_1(\omega\_{X/k}^\vee) \right)^n\_X = \frac{(a\_0+\dots+a\_n)^n}{a\_0\cdots a\_n}.$$


