Which Theorem to Use?

Select a scenario, make your initial guess, work the decision tree, then commit — the theorem cards reveal.

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Select a scenario, work through all tree steps, then press Commit answer to reveal the cards with your results.

One theorem beneath all five
\(\displaystyle\int_{\partial\Omega}\omega = \int_{\Omega}d\omega\)
Every theorem below is a special case of the Generalized Stokes' Theorem. See the Concept Connections sidebar.

Fundamental Theorem for Line Integrals

1-D analogue
Reduces a conservative line integral to two endpoint evaluations.
Requires \(\mathbf{F}=\nabla f\) and smooth oriented curve \(C\) from \(A\) to \(B\).

Let \(C\) be smooth from \(A\) to \(B\), \(\mathbf{F}=\nabla f\), \(\nabla f\) continuous on an open set containing \(C\).

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A)\]

Symbols

  • \(C\)Smooth oriented curve from \(A\) to \(B\)
  • \(\mathbf{F}=\nabla f\)Conservative: F must equal the gradient of a scalar potential \(f\)
  • \(d\mathbf{r}\)Vector line-element \(\langle dx,dy,dz\rangle\)
  • \(A,B\)Initial and terminal endpoints of \(C\)
Setup. Evaluate \(\int_C \mathbf{F}\cdot d\mathbf{r}\), \(\mathbf{F}=\langle 2xy+z^2,x^2,2xz\rangle\), any smooth curve from \((0,0,0)\) to \((1,2,3)\).
Verify conservative
Check all mixed partials: \(\partial_y(2xy+z^2)=2x=\partial_x(x^2)\), \(\partial_z(2xy+z^2)=2z=\partial_x(2xz)\), \(\partial_z(x^2)=0=\partial_y(2xz)\). All match. ✓
Find potential
Integrate: \(f=x^2y+xz^2\). Verify \(\nabla f=\mathbf{F}\). ✓
Apply FTC
\[\int_C\mathbf{F}\cdot d\mathbf{r}=f(1,2,3)-f(0,0,0)=(2+9)-0=11.\]
Answer
\(11\)
Path independence: the curve shape is irrelevant — only the endpoints enter the calculation.

Green's Theorem

Planar / 2-D
Converts a closed line integral over \(\partial D\) to a double integral of scalar curl over \(D\).
Requires \(D\subset\mathbb{R}^2\) simply connected, \(\partial D\) positively (CCW) oriented, \(P,Q\in C^1\).

Let \(D\subset\mathbb{R}^2\) be simply connected, \(C=\partial D\) positively (CCW) oriented, \(P,Q\in C^1\).

\[\oint_C P\,dx+Q\,dy=\iint_D\!\Bigl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigr)dA\]

Symbols

  • \(D\)Simply connected plane region in \(\mathbb{R}^2\)
  • \(C=\partial D\)Positively (CCW) oriented closed boundary
  • \(\tfrac{\partial Q}{\partial x}-\tfrac{\partial P}{\partial y}\)Scalar (2-D) curl of \(\mathbf{F}=\langle P,Q\rangle\)
Setup. Evaluate \(\oint_C(x^3-y^3)\,dx+(x^3+y^3)\,dy\), \(C\) the circle \(x^2+y^2=4\), CCW.
Scalar curl
\[\tfrac{\partial Q}{\partial x}-\tfrac{\partial P}{\partial y}=3x^2+3y^2=3r^2.\]
Polar integral
\[\iint_D 3r^2\,dA=2\pi\!\int_0^2 3r^3\,dr=24\pi.\]
Answer
\(24\pi\)
Degree-3 trig boundary expansions are replaced by a single polar double integral.

Stokes' Theorem

Surface / 3-D
Converts circulation around \(\partial S\) into a curl-flux integral over \(S\).
Requires oriented surface \(S\) in \(\mathbb{R}^3\), \(\partial S\) consistent with \(\hat{\mathbf{n}}\) by right-hand rule.

Let \(S\) be oriented with normal \(\hat{\mathbf{n}}\) and closed boundary curve \(C=\partial S\). Let \(\mathbf{F}\in C^1\).

\[\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}\]

Symbols

  • \(S\)Oriented smooth surface in \(\mathbb{R}^3\)
  • \(\hat{\mathbf{n}}\)Unit normal; fixes orientation of \(\partial S\) via right-hand rule
  • \(\nabla\times\mathbf{F}\)Curl of \(\mathbf{F}\)
  • \(d\mathbf{S}\)Vector surface element \(=\hat{\mathbf{n}}\,dS\)
Setup. Evaluate \(\oint_C \mathbf{F}\cdot d\mathbf{r}\), \(\mathbf{F}=\langle -y^2,x,z^2\rangle\), \(C\) = cylinder \(x^2+y^2=1\) ∩ plane \(z=2\), CCW from above.
Choose surface
Cap with disk \(S\): \(x^2+y^2\le 1\) in \(z=2\), upward \(\hat{k}\). Right-hand rule: CCW ↔ upward. ✓
Compute curl
\[\nabla\times\mathbf{F}=\langle 0,0,1+2y\rangle.\]
Apply Stokes' + symmetry
\[\oint_C=\iint_D(1+2y)\,dA=\pi+0=\pi.\]
Answer
\(\pi\)
Stokes' replaced a 3-D curve integral with a flat disk integral; odd-function symmetry kills \(\iint 2y\,dA\) immediately.

Divergence Theorem (Gauss)

Volume / 3-D
Converts outward flux through closed \(\partial E\) into a triple integral of divergence over \(E\).
Requires simple solid \(E\), piecewise-smooth closed \(\partial E\) with outward normal, \(\mathbf{F}\in C^1\).

Let \(E\) be a simple solid bounded by outward-oriented closed surface \(S=\partial E\). Let \(\mathbf{F}\in C^1\).

\[\oiint_S\!\mathbf{F}\cdot d\mathbf{S}=\iiint_E \nabla\cdot\mathbf{F}\;dV\]

Symbols

  • \(E\)Bounded solid region in \(\mathbb{R}^3\)
  • \(S=\partial E\)Closed surface with outward unit normal
  • \(\nabla\cdot\mathbf{F}\)Divergence: \(\partial_xF_1+\partial_yF_2+\partial_zF_3\)
  • \(dV\)Volume element inside \(E\)
Setup. Compute \(\oiint_S \mathbf{F}\cdot d\mathbf{S}\), \(\mathbf{F}=\langle x^3,y^3,z^3\rangle\), \(S\) the sphere \(\rho=a\) with outward normal.
Divergence
\[\nabla\cdot\mathbf{F}=3\rho^2.\]
Spherical integral
\[\iiint_E 3\rho^2\,dV=\tfrac{12\pi a^5}{5}.\]
Answer
\(\dfrac{12\pi a^5}{5}\)
The Divergence Theorem replaced six surface integrals with one spherical triple integral.

Surface Integrals (scalar & flux)

Direct / Fallback
Converts curved-surface integration to a flat \(D_{uv}\) integral via \(\|\mathbf{r}_u\times\mathbf{r}_v\|\).
Requires smooth parametrization \(\mathbf{r}(u,v)\) on \(D_{uv}\) with \(\mathbf{r}_u\times\mathbf{r}_v\neq\mathbf{0}\).
This is always available. Unlike Theorems 1–4, it requires no special structure — it is the direct method when no boundary theorem applies.

Let \(S=\mathbf{r}(D_{uv})\), \(\mathbf{r}_u=\partial_u\mathbf{r}\), \(\mathbf{r}_v=\partial_v\mathbf{r}\).

\[\iint_S f\,dS=\iint_{D_{uv}}f(\mathbf{r})\|\mathbf{r}_u\times\mathbf{r}_v\|\,dA\] \[\iint_S\mathbf{F}\cdot d\mathbf{S}=\iint_{D_{uv}}\mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,dA\]

Symbols

  • \(\mathbf{r}(u,v)\)Smooth parametrization of \(S\)
  • \(\mathbf{r}_u\times\mathbf{r}_v\)Normal vector; magnitude = area Jacobian
  • \(d\mathbf{S}\)Vector surface element \(=(\mathbf{r}_u\times\mathbf{r}_v)\,dA\)
Setup. Compute \(\iint_S z\,dS\), \(S\) the hemisphere \(x^2+y^2+z^2=4\), \(z\ge 0\). (Scenario F — scalar integrand.)
Spherical parametrize
\(\mathbf{r}(\phi,\theta)=\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\rangle\), \(0\le\phi\le\pi/2\), \(0\le\theta\le 2\pi\).
Jacobian
\[\|\mathbf{r}_\phi\times\mathbf{r}_\theta\|=4\sin\phi.\]
Integrate
On \(S\), \(z=2\cos\phi\).
\[\iint_S z\,dS=\int_0^{2\pi}\!\int_0^{\pi/2}(2\cos\phi)(4\sin\phi)\,d\phi\,d\theta=8\pi.\]
Answer
\(8\pi\)
The Jacobian \(4\sin\phi\) accounts for the area-stretching from pole to equator — without it, you'd compute an ordinary double integral over \((\phi,\theta)\) space instead of actual surface area.

Line Integral (direct parametrization)

Direct / Fallback
Converts a path integral to an ordinary single-variable integral via a parametrization \(\mathbf{r}(t)\).
Use when \(\mathbf{F}\) is not conservative, the curve is not closed, or no theorem shortcut applies.
This always works for curve integrals. It is the fallback when FTC, Green's, and Stokes' do not apply.

Let \(C\) be parametrized by \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), \(a\le t\le b\), \(\mathbf{r}'(t)\) continuous.

\[\int_C\mathbf{F}\cdot d\mathbf{r}=\int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt\] \[\int_C f\,ds=\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\]

Symbols

  • \(\mathbf{r}(t)\)Smooth parametrization of the curve \(C\), \(a\le t\le b\)
  • \(\mathbf{r}'(t)\)Tangent vector to \(C\); its magnitude \(\|\mathbf{r}'(t)\|\) is the arc-length scaling factor
  • \(d\mathbf{r}\)\(=\mathbf{r}'(t)\,dt\); the directed line element
  • \(ds\)\(=\|\mathbf{r}'(t)\|\,dt\); the scalar arc-length element
Setup. Evaluate \(\int_C\mathbf{F}\cdot d\mathbf{r}\), \(\mathbf{F}=\langle y^2,x\rangle\), \(C\) the parabola \(y=x^2\) from \((0,0)\) to \((1,1)\).
Confirm not conservative
\(\partial_y(y^2)=2y\neq 1=\partial_x(x)\). Mixed partials differ — FTC does not apply.
Parametrize
\(\mathbf{r}(t)=\langle t,t^2\rangle\), \(0\le t\le 1\), \(\mathbf{r}'(t)=\langle 1,2t\rangle\).
Substitute
\(\mathbf{F}(\mathbf{r}(t))=\langle t^4,t\rangle\).
\[\mathbf{F}\cdot\mathbf{r}'=t^4(1)+t(2t)=t^4+2t^2.\]
Integrate
\[\int_0^1(t^4+2t^2)\,dt=\tfrac{1}{5}+\tfrac{2}{3}=\tfrac{13}{15}.\]
Answer
\(\dfrac{13}{15}\)
Because \(\mathbf{F}\) is not conservative, the answer depends on the specific path — a different curve from \((0,0)\) to \((1,1)\) gives a different value.