= Solution
Take a fixed <sphere> of radius $R$ in an unbounded incompressible inviscid fluid, with uniform <velocity> $U\mathbf e_z$ at infinity. Assume <irrotational flow>, so $\mathbf v=\nabla\phi$ and <incompressibility> gives $\nabla^2\phi=0$. Axisymmetry and the dipolar angular dependence of the far field suggest
$$
\phi=(Ar+B/r^2)\cos\theta.
$$
Matching the far field gives $A=U$. The no-penetration condition $v_r(R)=0$ then gives $B=UR^3/2$. Thus
$$
\boxed{\phi=U\left(r+\frac{R^3}{2r^2}\right)\cos\theta,}
$$
$$
v_r=U\left(1-\frac{R^3}{r^3}\right)\cos\theta,\qquad
v_\theta=-U\left(1+\frac{R^3}{2r^3}\right)\sin\theta.
$$
The surface has <stagnation points> at its front and rear poles, while the equatorial speed is $3U/2$. Fluid slips tangentially past the <sphere>; imposing a no-slip condition would contradict this inviscid potential-flow model.
With gravity omitted or the hydrostatic <pressure> subtracted, steady Bernoulli gives
$$
p+\frac12\rho v^2=p_\infty+\frac12\rho U^2.
$$
At the surface,
$$
\boxed{p(R,\theta)=p_\infty+\frac12\rho U^2
\left(1-\frac94\sin^2\theta\right).}
$$
The front–rear symmetry makes the <integral> of <pressure> <force> along the flow vanish: terms proportional to $\cos\theta$ integrate to zero over the <sphere>. This is <D'Alembert's paradox>, not a prediction that real spheres have no <drag>. <Viscosity> creates <boundary layers>; separation and a wake break the fore–aft symmetry. <Potential flow> remains valuable for the outer <velocity> field when viscous regions are thin, but it cannot supply their <drag> by itself.
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