Coverage: Comprehensive — all course material (Lessons 1–37)
Exam Date: Tuesday, May 05, 2026
Time: 1:00 PM - 3:00 PM
Location: Announced on Brightspace
Stokes' Theorem [Lessons 34–35] (§17.7)
Stokes' Theorem
Let \(S\) be an oriented, piecewise-smooth surface bounded by a simple closed curve \(C = \partial S\). Let \(\vec{F} = \langle P, Q, R \rangle\) have continuous partial derivatives on an open region containing \(S\). Then:
Equivalently, using \(d\vec{S} = \hat{n}\,dS\):
Orientation Convention
The boundary curve \(C\) is positively oriented relative to \(S\) when, walking along \(C\) with the surface on your left, \(\hat{n}\) points away from your head. Equivalently: \(\hat{n}\) and the direction of traversal of \(C\) satisfy the right-hand rule.
Curl Formula
For \(\vec{F} = \langle P, Q, R \rangle\):
Computing the Surface Integral Side
For surface \(z = g(x,y)\) with upward normal over region \(D\):
Key: Stokes' theorem converts a line integral into a surface integral (or vice versa). It generalizes Green's Theorem: when \(S\) is flat in the \(xy\)-plane, Stokes' theorem reduces exactly to Green's theorem.
Connection to Green's Theorem
Green's Theorem is the special case of Stokes' Theorem where \(S\) lies in the \(xy\)-plane (\(z=0\), \(\hat{n}=\vec{k}\)):
Consequences
- If \(\operatorname{curl}(\vec{F}) = \vec{0}\) (i.e., \(\vec{F}\) is conservative), then \(\oint_C \vec{F}\cdot d\vec{r} = 0\) for every closed curve \(C\).
- Two surfaces \(S_1\) and \(S_2\) with the same oriented boundary \(C\) give the same surface integral of \(\operatorname{curl}(\vec{F})\).
- Stokes' theorem can simplify computation by replacing a hard surface with an easier one sharing the same boundary.
Practice Problems
Problem 1. Use Stokes' Theorem to evaluate \(\displaystyle\oint_C \vec{F}\cdot d\vec{r}\) where \(\vec{F}=\langle y^2, x, z^2\rangle\) and \(C\) is the boundary of the triangle with vertices \((1,0,0)\), \((0,1,0)\), \((0,0,1)\), oriented counterclockwise when viewed from above.
▼ Show answerProblem 2. Let \(\vec{F}=\langle -y^2, x, z^2\rangle\). Use Stokes' Theorem to evaluate \(\displaystyle\oint_C\vec{F}\cdot d\vec{r}\) where \(C\) is the circle \(x^2+y^2=4\) in the plane \(z=3\), traversed counterclockwise when viewed from above.
▼ Show answerProblem 3. Evaluate \(\displaystyle\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S}\) where \(\vec{F}=\langle yz, xz, xy\rangle\) and \(S\) is the part of the sphere \(x^2+y^2+z^2=4\) with \(z\ge0\), oriented with upward normal. (Hint: use Stokes' Theorem and evaluate the line integral on \(\partial S\) instead.)
▼ Show answerProblem 4. Use Stokes' Theorem to evaluate \(\displaystyle\oint_C \vec{F}\cdot d\vec{r}\) where \(\vec{F}=\langle xz, yz, xy\rangle\) and \(C\) is the boundary of the surface \(z = 1 - x^2 - y^2\) above the \(xy\)-plane, traversed counterclockwise when viewed from above.
▼ Show answerProblem 5. Let \(S\) be the part of the paraboloid \(z=9-x^2-y^2\) above \(z=0\) with upward normal, and \(\vec{F}=\langle -y^3, x^3, -z^3\rangle\). Use Stokes' Theorem to find \(\displaystyle\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S}\).
▼ Show answerDivergence Theorem [Lessons 36–37] (§17.9)
Divergence Theorem (Gauss's Theorem)
Let \(E\) be a simple solid region with outward-oriented boundary surface \(S = \partial E\). Let \(\vec{F} = \langle P, Q, R \rangle\) have continuous partial derivatives on an open region containing \(E\). Then:
Divergence Formula
Orientation Convention
The boundary surface \(S = \partial E\) must be oriented with the outward normal (pointing away from the interior of \(E\)). If the surface is given with inward normal, multiply the result by \(-1\).
Key: The Divergence Theorem converts a flux surface integral into a triple integral over the enclosed volume — often much easier to evaluate. It is especially useful when the surface consists of multiple pieces.
Physical Interpretation
The Divergence Theorem states that the net flux of \(\vec{F}\) out of a closed surface equals the total divergence (source/sink density) integrated over the enclosed volume. If \(\operatorname{div}(\vec{F}) = 0\) everywhere in \(E\), then the net flux through \(\partial E\) is zero.
Strategy: Replacing a Hard Surface
If \(S\) is an open surface (not closed), close it by adding a cap \(S_{\text{cap}}\), apply the Divergence Theorem to the closed surface \(S \cup S_{\text{cap}}\), then subtract the cap's contribution:
Key Identities
- \(\operatorname{div}(\operatorname{curl}(\vec{F})) = 0\) for any smooth \(\vec{F}\) — so \(\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S} = 0\) for any closed \(S\)
- If \(\operatorname{div}(\vec{F}) = 0\) in \(E\), the net outward flux through \(\partial E\) is zero
- Volume of \(E\): \(V = \frac{1}{3}\iint_S \vec{r}\cdot\hat{n}\,dS = \frac{1}{3}\iint_S \langle x,y,z\rangle\cdot d\vec{S}\)
Practice Problems
Problem 1. Use the Divergence Theorem to evaluate \(\displaystyle\iint_S \vec{F}\cdot d\vec{S}\) where \(\vec{F}=\langle x^3, y^3, z^3\rangle\) and \(S\) is the sphere \(x^2+y^2+z^2=1\) with outward normal.
▼ Show answerProblem 2. Use the Divergence Theorem to compute the flux of \(\vec{F}=\langle xy, y^2, yz\rangle\) outward through the surface of the unit cube \([0,1]^3\).
▼ Show answerProblem 3. Let \(S\) be the closed surface consisting of the hemisphere \(z=\sqrt{4-x^2-y^2}\) and the disk \(x^2+y^2\le4\) in the \(xy\)-plane, both with outward normal. Evaluate \(\displaystyle\iint_S \vec{F}\cdot d\vec{S}\) where \(\vec{F}=\langle x,y,z\rangle\).
▼ Show answerProblem 4. Use the Divergence Theorem to find the outward flux of \(\vec{F}=\langle x^2z, y^2z, z^2\rangle\) through the boundary of the solid region \(E = \{(x,y,z)\mid x^2+y^2\le4,\;0\le z\le3\}\) (a cylinder).
▼ Show answerProblem 5. Let \(S\) be the outward-oriented boundary of the region between the spheres \(x^2+y^2+z^2=1\) and \(x^2+y^2+z^2=4\). Evaluate \(\displaystyle\iint_S \vec{F}\cdot d\vec{S}\) where \(\vec{F}=\langle x, y, z\rangle\).
▼ Show answer