Exercise 1520Calculus › Multivariable calculus › Vector calculusL3Added 2 Nov 2026

Stokes' theorem

#Stokes#curl#surface integral

Statement

Let $S$ be the planar surface $z = 0$ bounded by the circle $x^2+y^2 = 1$ , and $C$ its oriented boundary. Compute $\int_C F \cdot dr$ for $F(x,y,z) = (-y, x, 0)$ .

Hint

Apply Stokes: $\int_C F \cdot dr = \iint_S (\nabla \times F) \cdot n \, dS$ .

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Solution

Curl.
$\nabla \times F = \left( \frac{\partial}{\partial y}(0) - \frac{\partial}{\partial z}(x), \frac{\partial}{\partial z}(-y) - \frac{\partial}{\partial x}(0), \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y) \right) = (0, 0, 2).$

Normal. $n = (0,0,1)$ for the upward-oriented surface.

Stokes.
$\int_C F \cdot dr = \iint_S 2 \, dS = 2 \cdot \text{Area}(S) = 2\pi.$

Remark. Directly, $F$ along the unit circle is $(-y,x,0)$ , and $\int_C (-y dx + x dy) = \int_0^{2\pi} (\sin^2 t + \cos^2 t)dt = 2\pi$ , confirming the result.

Stokes' theorem

Statement

Solution

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