Volume of $-K_X$ for a weighted projective variety 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 
$D = \sum_{i=0}^n c_i D_i$? 
 A: 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, $$\left( c_1(\omega_{X/k}^\vee) \right)^n_X = \frac{(a_0+\dots+a_n)^n}{a_0\cdots a_n}.$$
