Let \(S\) be a smooth surface with a smooth bounding curve \(C.\) Then for any continuously differentiable vector function

\[{\mathbf{F}\left( {x,y,z} \right) \text{ = }}\kern0pt{ \big( {P\left( {x,y,z} \right),\,}}\kern0pt{{Q\left( {x,y,z} \right),\,}}\kern0pt{{R\left( {x,y,z} \right)} \big)}\]

Stoke’s Theorem states:

\[{\oint\limits_C {\mathbf{F} \cdot d\mathbf{r}} }={ \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} ,}\]

where

\[ {\nabla \times \mathbf{F} } = {\left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ P&Q&R \end{array}} \right| } = {\left( {\frac{{\partial R}}{{\partial y}} – \frac{{\partial Q}}{{\partial z}}} \right)\mathbf{i} } + {\left( {\frac{{\partial P}}{{\partial z}} – \frac{{\partial R}}{{\partial x}}} \right)\mathbf{j} } + {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right)\mathbf{k} } \]

is the curl of the vector field \(\mathbf{F}.\)

The symbol \(\oint\limits{}\) indicates that the line integral is taken over a closed curve.

We assume there is an orientation on both the surface and the curve that are related by the right hand rule. That is, if you were to walk around the curve in its preferred direction with your head pointing in the same direction as the normal vector \(\mathbf{n}\) to the surface, then the surface would always be on your left (Figure \(1\)).

Stoke’s Theorem relates line integrals of vector fields to surface integrals of vector fields.

In coordinate form Stoke’s Theorem can be written as

\[

{\oint\limits_C {Pdx + Qdy + Rdz} }

= {\iint\limits_S {\left( {\frac{{\partial R}}{{\partial y}} – \frac{{\partial Q}}{{\partial z}}} \right)dydz }}+{{ \left( {\frac{{\partial P}}{{\partial z}} – \frac{{\partial R}}{{\partial x}}} \right)dzdx }}+{{ \left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right)dxdy} .}

\]

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Show that the line integral \(\oint\limits_C {yzdx }\) \({+\, xzdy }\) \({+\, xydz} \) is zero along any closed contour \(C.\)### Example 2

Use Stoke’s Theorem to evaluate the line integral \(\oint\limits_C {\left( {y + 2z} \right)dx }\) \({+ \left( {x + 2z} \right)dy }\) \({+ \left( {x + 2y} \right)dz},\) where \(C\) is the curve formed by intersection of the sphere \({x^2} + {y^2} + {z^2} \) \(= 1\) with the plane \(x + 2y + 2z \) \(= 0.\)### Example 3

Use Stoke’s Theorem to calculate the line integral \(\oint\limits_C {{y^3}dx – {x^3}dy }\) \({+\, {z^3}dz}.\) The curve \(C\) is the intersection of the cylinder \({x^2} + {y^2} = {a^2}\) and the plane \(x + y + z \) \(= b.\)### Example 4

Use Stoke’s Theorem to evaluate the line integral \(\oint\limits_C {\left( {x + z} \right)dx }\) \({+ \left( {x – y} \right)dy }\) \({+\, xdz}.\) The curve \(C\) is the ellipse defined by the equation \({\large\frac{{{x^2}}}{4}\normalsize} + {\large\frac{{{y^2}}}{9}\normalsize} = 1,\) \(z = 1\) (Figure \(2\)).### Example 5

Use Stoke’s Theorem to calculate the line integral \(\oint\limits_C {\left( {z – y} \right)dx }\) \({+ \left( {x – z} \right)dy }\) \({+ \left( {y – x} \right)dz} .\) The curve \(C\) is the triangle with the vertices \(A\left( {2,0,0} \right),\) \(B\left( {0,2,0} \right),\) \(D\left( {0,0,2} \right)\) (Figure \(3\)).### Example 6

Use Stoke’s Theorem to evaluate the line integral \(\oint\limits_C {\left( {{z^2} – {y^2}} \right)dx }\) \({+\, \left( {{x^2} – {z^2}} \right)dy }\) \({+\, \left( {{y^2} – {x^2}} \right)dz},\) where the curve \(C\) is formed by intersection of the paraboloid \(z =\) \(5 – {x^2} – {y^2}\) with the plane \(x + y + z \) \(= 1\) (Figure \(4\)).### Example 1.

Show that the line integral \(\oint\limits_C {yzdx }\) \({+\, xzdy }\) \({+\, xydz} \) is zero along any closed contour \(C.\)Solution.

Let \(S\) be a surface bounded by a closed curve \(C.\) Applying Stoke’s formula, we identify that

\[{P = yz,\;\;\;}\kern-0.3pt{Q = xz,\;\;\;}\kern-0.3pt{R = xy.}\]

Then

\[ {\nabla \times {\mathbf{F}} } = {\left( {\frac{{\partial R}}{{\partial y}} – \frac{{\partial Q}}{{\partial z}}} \right){\mathbf{i}} } + {\left( {\frac{{\partial P}}{{\partial z}} – \frac{{\partial R}}{{\partial x}}} \right){\mathbf{j}} } + {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right){\mathbf{k}} } = {\left( {x – x} \right)\mathbf{i} }+{ \left( {y – y} \right)\mathbf{j} }+{ \left( {z – z} \right)\mathbf{k} } = {0 \cdot \mathbf{i} }+{ 0 \cdot \mathbf{j} }+{ 0 \cdot \mathbf{k} }={ \mathbf{0}.} \]

Hence, we can find the line integral:

\[

{\oint\limits_C {yzdx + xzdy + xydz} }

= {\iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} }

= {\iint\limits_S {\mathbf{0} \cdot d\mathbf{S}} = 0.}

\]

So the statement is proved.

### Example 2.

Use Stoke’s Theorem to evaluate the line integral \(\oint\limits_C {\left( {y + 2z} \right)dx }\) \({+ \left( {x + 2z} \right)dy }\) \({+ \left( {x + 2y} \right)dz},\) where \(C\) is the curve formed by intersection of the sphere \({x^2} + {y^2} + {z^2} \) \(= 1\) with the plane \(x + 2y + 2z \) \(= 0.\)Solution.

Let \(S\) be the circle cut by the sphere from the plane. Find the coordinates of the unit vector \(\mathbf{n}\) normal to the surface \(S:\)

\[

{\mathbf{n} }={ \frac{{1 \cdot \mathbf{i} + 2 \cdot \mathbf{j} + 2 \cdot \mathbf{k}}}{{\sqrt {{1^2} + {2^2} + {2^2}} }} }

= {\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}.}

\]

In our case

\[{P = y + 2z,\;\;\;}\kern-0.3pt{Q = x + 2z,\;\;\;}\kern-0.3pt{R = x + 2y.}\]

Hence, the curl of the vector \(\mathbf{F}\) is

\[ {\nabla \times {\mathbf{F}} } = {\left( {\frac{{\partial R}}{{\partial y}} – \frac{{\partial Q}}{{\partial z}}} \right){\mathbf{i}} } + {\left( {\frac{{\partial P}}{{\partial z}} – \frac{{\partial R}}{{\partial x}}} \right){\mathbf{j}} } + {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right){\mathbf{k}} } = {\left( {2 – 2} \right){\mathbf{i}} }+{ \left( {2 – 1} \right){\mathbf{j}} }+{ \left( {1 – 1} \right){\mathbf{k}} } = {{\bf{j}}.} \]

Using Stoke’s Theorem, we have

\[

{\oint\limits_C {\left( {y + 2z} \right)dx + \left( {x + 2z} \right)dy }+{ \left( {x + 2y} \right)dz} }

= {\iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} }

= {\iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot \mathbf{n}dS} }

= {\iint\limits_S {\mathbf{j} \cdot \left( {\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}} \right)dS} }

= {\frac{2}{3}\iint\limits_S {dS} .}

\]

As the sphere \({x^2} + {y^2} + {z^2} \) \(= 1\) is centered at the origin and the plane \(x + 2y + 2z \) \(= 0\) also passes through the origin, the cross section is the circle of radius \(1.\) Hence the integral is

\[{I = \frac{2}{3}\iint\limits_S {dS} }={ \frac{2}{3} \cdot \pi \cdot {1^2} }={ \frac{{2\pi }}{3}.}\]