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.
Figure 2. 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 \(3.\)
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
\[{S_{ABD}} = \frac{1}{2}\left| {\overrightarrow{AB} \times \overrightarrow{BD}} \right| = \frac{1}{2} \cdot 4\sqrt 3 = 2\sqrt 3 .\]
The complete answer is
\[I = 2\sqrt 3 \iint\limits_S {dS} = 2\sqrt 3 \cdot 2\sqrt 3 = 12.\]
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 \(4.\)
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
\[I = 4 \cdot \frac{{9\pi }}{2} = 18\pi .\]
