# Stoke’s Theorem

## Solved Problems

### 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 $$x + y + z = b.$$

Solution.

We suppose that $$S$$ is the part of the plane cut by the cylinder. The curve $$C$$ is oriented counterclockwise when viewed from the end of the normal vector $$\mathbf{n},$$ which has coordinates

$\mathbf{n} = \frac{{1 \cdot \mathbf{i} + 1 \cdot \mathbf{j} + 1 \cdot \mathbf{k}}}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \frac{1}{{\sqrt 3 }}\mathbf{i} + \frac{1}{{\sqrt 3 }}\mathbf{j} + \frac{1}{{\sqrt 3 }}\mathbf{k}.$

As $$P = {y^3},$$ $$Q = -{x^3},$$ $$R = {z^3},$$ we can write:

$\nabla \times {\mathbf{F}} = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right){\mathbf{i}} + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right){\mathbf{j}} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right){\mathbf{k}} = \left( {0 - 0} \right){\mathbf{i}} + \left( {0 - 0} \right){\mathbf{j}} + \left( { - 3{x^2} - 3{y^2}} \right){\mathbf{k}} = - 3\left( {{x^2} + {y^2}} \right)\mathbf{k}.$

Applying Stoke's Theorem, we find:

$I = \oint\limits_C {{y^3}dx - {x^3}dy + {z^3}dz} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot \mathbf{n}dS} = \iint\limits_S {\left( { - 3\left( {{x^2} + {y^2}} \right)\mathbf{k}} \right) \cdot \left( {\frac{1}{{\sqrt 3 }}\mathbf{i} + \frac{1}{{\sqrt 3 }}\mathbf{j} + \frac{1}{{\sqrt 3 }}\mathbf{k}} \right)dS} = - \sqrt 3 \iint\limits_S {\left( {{x^2} + {y^2}} \right)dS}.$

We can express the surface integral in terms of the double integral:

$I = - \sqrt 3 \iint\limits_S {\left( {{x^2} + {y^2}} \right)dS} = - \sqrt 3 \iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right) \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} .$

The equation of the plane is $$z = b - x - y,$$ so the square root in the integrand is equal to

$\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} = \sqrt {1 + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt 3 .$

Hence,

$I = - \sqrt 3 \iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right)\sqrt 3 dxdy} = - 3\iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right)dxdy} .$

The region $$D\left( {x,y} \right)$$ is the circle $${x^2} + {y^2} = {a^2}$$ of radius $$a.$$ By changing to polar coordinates, we get

$I = - 3\int\limits_0^{2\pi } {\int\limits_0^a {{r^3}drd\theta } } = - 3 \cdot 2\pi \cdot \left. {\frac{{{r^4}}}{4}} \right|_0^a = - \frac{{3\pi {a^4}}}{2}.$

### 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 $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1,$$ $$z = 1$$ (Figure $$2$$).

Solution.

Let the surface $$S$$ be the part of the plane $$z = 1$$ bounded by the ellipse. Obviously that the unit normal vector is $$\mathbf{n} = \mathbf{k}.$$ Since

$P = x + z,\;\; Q = x - y,\;\; R = x,$

then the curl of the vector field $$\mathbf{F}$$ is

$\nabla \times {\mathbf{F}} = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right){\mathbf{i}} + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right){\mathbf{j}} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right){\mathbf{k}} = \left( {0 - 0} \right){\mathbf{i}} + \left( {1 - 1} \right){\mathbf{j}} + \left( {1 - 0} \right){\mathbf{k}} = \mathbf{k}.$

By Stoke's Theorem,

$\oint\limits_C {\left( {x + z} \right)dx + \left( {x - y} \right)dy + xdz} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot \mathbf{n}dS} = \iint\limits_S {\mathbf{k} \cdot \mathbf{k}dS} = \iint\limits_S {dS} .$

The double integral in the latter formula is the area of the ellipse. Therefore, the integral is

$\iint\limits_S {dS} = \pi \cdot 2 \cdot 3 = 6\pi .$

### 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 $$3$$).

Solution.

We suppose that the surface $$S$$ is the plane of the triangle $$ABD.$$ Orientation of the surface $$S$$ and the contour $$C$$ are shown in Figure $$3.$$

We first find the unit normal vector $$\mathbf{n}:$$

$\overrightarrow{AB} = \big( {{x_B} - {x_A},\; {y_B} - {y_A},\; {z_B} - {z_A}} \big) = \left( {0 - 2,2 - 0,0 - 0} \right) = \left( { - 2,2,0} \right),$
$\overrightarrow{BD} = \big( {{x_D} - {x_B},\; {y_D} - {y_B},\; {z_D} - {z_B}} \big) = \left( {0 - 0,0 - 2,2 - 0} \right) = \left( { 0,-2,2} \right).$

Then

$\overrightarrow{AB} \times \overrightarrow{BD} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ { - 2} & 2 & 0\\ 0 & { - 2} & 2 \end{array}} \right| = 4\mathbf{i} + 4\mathbf{j} + 4\mathbf{k},$

and hence,

$\mathbf{n} = \frac{{\overrightarrow{AB} \times \overrightarrow{BD}}}{{\left| {\overrightarrow{AB} \times \overrightarrow{BD}} \right|}} = \frac{{4\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}}}{{\sqrt {{4^2} + {4^2} + {4^2}} }} = \frac{1}{{\sqrt 3 }}\mathbf{i} + \frac{1}{{\sqrt 3 }}\mathbf{j} + \frac{1}{{\sqrt 3 }}\mathbf{k}.$

In our case $$P = z - y,$$ $$Q = x - z,$$ $$R = y - x,$$ so the curl of $$\mathbf{F}$$ is

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

By Stoke's formula,

$I = \oint\limits_C {\left( {z - y} \right)dx + \left( {x - z} \right)dy }+{ \left( {y - x} \right)dz} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot \mathbf{n}dS} = \iint\limits_S {\left( {2{\mathbf{i}} + 2{\mathbf{j}} + 2{\mathbf{k}}} \right) \cdot \left( {\frac{1}{{\sqrt 3 }}\mathbf{i} + \frac{1}{{\sqrt 3 }}\mathbf{j} + \frac{1}{{\sqrt 3 }}\mathbf{k}} \right)dS} = \frac{2}{{\sqrt 3 }}\iint\limits_S {\left( {1 + 1 + 1} \right)dS} = 2\sqrt 3 \iint\limits_S {dS} .$

Here the double integral $$\iint\limits_S {dS}$$ is the area of the triangle $$ABD,$$ which is equal to

${S_{ABD}} = \frac{1}{2}\left| {\overrightarrow{AB} \times \overrightarrow{BD}} \right| = \frac{1}{2} \cdot 4\sqrt 3 = 2\sqrt 3 .$

$I = 2\sqrt 3 \iint\limits_S {dS} = 2\sqrt 3 \cdot 2\sqrt 3 = 12.$

### 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 $$x + y + z = 1.$$

Solution.

Let $$S$$ be the part of the plane cut by the paraboloid. Orientation of the surface $$S$$ and the curve $$C$$ are shown in Figure $$4.$$

The normal vector $$\mathbf{n}$$ can be found from the equation of the plane:

$\mathbf{n} = \frac{{1 \cdot \mathbf{i} + 1 \cdot \mathbf{j} + 1 \cdot \mathbf{k}}}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \frac{1}{{\sqrt 3 }}\mathbf{i} + \frac{1}{{\sqrt 3 }}\mathbf{j} + \frac{1}{{\sqrt 3 }}\mathbf{k}.$

Since

$P = {z^2} - {y^2},\;\; Q = {x^2} - {z^2},\;\; R = {y^2} - {x^2},$

the curl of the vector field $$\mathbf{F}$$ is

$\nabla \times {\mathbf{F}} = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right){\mathbf{i}} + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right){\mathbf{j}} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right){\mathbf{k}} = \left( {2y + 2z} \right){\mathbf{i}} + \left( {2z + 2x} \right){\mathbf{j}} + \left( {2x + 2y} \right){\mathbf{k}}.$

By Stoke's formula, we have

$I = \oint\limits_C {\left( {{z^2} - {y^2}} \right)dx + \left( {{x^2} - {z^2}} \right)dy + \left( {{y^2} - {x^2}} \right)dz} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot d\mathbf{S}} = \iint\limits_S {\left( {\nabla \times \mathbf{F}} \right) \cdot \mathbf{n}dS} = \frac{2}{{\sqrt 3 }}\iint\limits_S {\left( {y + z + z + x + x + y} \right)dS} = \frac{4}{{\sqrt 3 }}\iint\limits_S {\left( {x + y + z} \right)dS} .$

As $$x + y + z = 1,$$ the integral becomes

$I = \frac{4}{{\sqrt 3 }}\iint\limits_S {dS} .$

To complete the calculation, we must evaluate the double integral $$\iint\limits_S {dS},$$ i.e. the area of the surface $$S.$$ The explicit equation of the plane is $$z = 1 - x - y.$$ Therefore, using the formula

$\iint\limits_S {dS} = \iint\limits_{D\left( {x,y} \right)} {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy},$

where $$D\left( {x,y} \right)$$ is projection of $$S$$ onto the $$xy$$-plane, we have

$I = \frac{4}{{\sqrt 3 }}\iint\limits_S {dS} = \frac{4}{{\sqrt 3 }} \iint\limits_{D\left( {x,y} \right)} {\sqrt {1 + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} dxdy} = 4\iint\limits_{D\left( {x,y} \right)} {dxdy} .$

Determine the region of integration $$D\left( {x,y} \right).$$ Solving the system of the equations

$\left\{ \begin{array}{l} x + y + z = 1\\ z = 5 - {x^2} - {y^2} \end{array} \right.,$

we obtain

$x + y + 5 - {x^2} - {y^2} = 1,\;\; \Rightarrow {x^2} + {y^2} - x - y = 4,\;\; \Rightarrow \left( {{x^2} - x + \frac{1}{4}} \right) + \left( {{y^2} - y + \frac{1}{4}} \right) = \frac{9}{2},\;\; \Rightarrow \left( {x - \frac{1}{2}} \right)^2 + \left( {y - \frac{1}{2}} \right)^2 = \left( {\frac{3}{{\sqrt 2 }}} \right)^2.$

Thus, we see that the region $$D\left( {x,y} \right)$$ is the circle of radius $$R = {\frac{3}{{\sqrt 2 }}}$$ centered at $$\left( {{\frac{1}{2}}, {\frac{1}{2}}} \right).$$ Then the area of the region $$D\left( {x,y} \right)$$ is

$\iint\limits_{D\left( {x,y} \right)} {dxdy} = \pi {\left( {\frac{3}{{\sqrt 2 }}} \right)^2} = \frac{{9\pi }}{2}.$

Hence, the initial integral is

$I = 4 \cdot \frac{{9\pi }}{2} = 18\pi .$