C.7 · Coordinate changes and vector integral theorems

Changing variables scales area or volume by the absolute Jacobian determinant. Polar coordinates use r dr dtheta; cylindrical coordinates use r dr dtheta dz; spherical coordinates with phi measured from the positive z-axis use rho² sin(phi) dρ dphi dtheta. State angle conventions and transform both the integrand and the region. A missing Jacobian changes a uniform density integral into the wrong physical quantity.

Green's theorem equates positively oriented planar boundary circulation integral P dx+Q dy with the double integral of Q_x−P_y on the region. The usual hypotheses require first derivatives continuous on an open set containing the region. A hole needs its own negatively oriented inner boundary, or another valid treatment of the missing domain. The theorem cannot integrate across a field singularity.

Stokes' theorem equates circulation on a surface boundary with the surface integral of curl F dot the oriented normal. The right-hand rule links boundary direction to the normal. The divergence theorem equates outward flux across a closed surface with the volume integral of div F. Circulation, flux, curl and divergence are distinct; a closed surface is required for the usual divergence theorem.

A gradient field has path-independent line integrals, determined by endpoint potential differences. A continuously differentiable curl-free field on a simply connected open domain is conservative. Curl-free alone on a domain with a hole is insufficient. For F=(−y/(x²+y²),x/(x²+y²)), the origin is excluded; unit-circle circulation is 2π, although the curl is zero wherever the field is defined.

For F=(x,y,z), divergence is 3. The outward flux through the sphere of radius 2 is therefore 3 times its volume, or 3·(4π·2³/3)=32π. This avoids a surface parameterisation. For the planar field (−y,x), Green's theorem gives counterclockwise unit-circle circulation as integral of 1−(−1)=2 over the disk, hence 2π. Reversing the orientation changes the circulation sign.

Zero curl is not enough when the domain has a hole. Outward flux and counterclockwise circulation use different theorems and orientation rules.

Evaluate $\iint_D(x^2+y^2)\,dA$ on $1\le x^2+y^2\le4$. Find the outward flux of $F=(x,y,z)$ across the unit sphere.

Worked solution

The annulus has polar bounds $1\le r\le2$, $0\le\theta\le2\pi$. Including the Jacobian,

F is continuously differentiable on the whole ball and has divergence three. The divergence theorem gives outward flux $3(4\pi/3)=4\pi$. Inward orientation would reverse its sign.

Let $G=(-y/(x^2+y^2),x/(x^2+y^2))$ on the punctured plane. Its scalar curl is zero there. Evaluate its circulation on the counterclockwise unit circle. Explain why Green's theorem does not make it zero.

Check after attempting

Parameterise $r(t)=(\cos t,\sin t)$ for $0\le t\le2\pi$. Then $G(r(t))=(-\sin t,\cos t)=r'(t)$, so $G\cdot r'=1$ and the integral is $2\pi$. The field is undefined at the origin inside the disk. Green's theorem requires the needed smoothness on a neighbourhood of the region, not just on its boundary. Zero curl on a punctured domain does not establish a global potential; a closed nonexact field can circulate around its hole.