Symplectic blowup size changes volume ($\mathrm{Vol}(\tilde{X})=\mathrm{Vol}(X)-{\epsilon }^n/n!$)

For a closed $2n$-dimensional symplectic manifold, a symplectic blowup of size $\epsilon =\pi R^2$ removes the symplectic volume of the radius-$R$ ball. Its exceptional divisor has lower dimension and contributes no ambient volume. The displayed volume loss shows that different size choices can give forms that are not related by a symplectomorphism.