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    Home / Calculus III / Surface Integrals / Stokes' Theorem
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

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    August 27, 2020

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    Section 6-5 : Stokes' Theorem

    In this section we are going to take a look at a theorem that is a higher dimensional version of Green’s Theorem. In Green’s Theorem we related a line integral to a double integral over some region. In this section we are going to relate a line integral to a surface integral. However, before we give the theorem we first need to define the curve that we’re going to use in the line integral.

    Let’s start off with the following surface with the indicated orientation.

    This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis moves off the right and slightly downward.  This is the upper hemisphere without a specified radius and whose base is at an unspecified point on the positive z-axis.  At several places on hemisphere there are normal vectors pointing outwards from the hemisphere.  Along the base of the hemisphere is a curve, labeled C, with arrow heads indicated that it is traced out in a counter clockwise motion if viewed from above.

    Around the edge of this surface we have a curve \(C\). This curve is called the boundary curve. The orientation of the surface \(S\) will induce the positive orientation of \(C\). To get the positive orientation of \(C\) think of yourself as walking along the curve. While you are walking along the curve if your head is pointing in the same direction as the unit normal vectors while the surface is on the left then you are walking in the positive direction on \(C\).

    Now that we have this curve definition out of the way we can give Stokes’ Theorem.

    Stokes’ Theorem

    Let \(S\) be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve \(C\) with positive orientation. Also let \(\vec F\) be a vector field then,

    \[\int\limits_{C}{{\vec F\centerdot \,d\,\vec r}} = \iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\,\centerdot \,d\vec S}}\]

    In this theorem note that the surface \(S\) can actually be any surface so long as its boundary curve is given by \(C\). This is something that can be used to our advantage to simplify the surface integral on occasion.

    Let’s take a look at a couple of examples.

    Example 1 Use Stokes’ Theorem to evaluate \(\displaystyle \iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\,\centerdot \,d\vec S}}\) where \(\vec F = {z^2}\,\vec i - 3xy\,\vec j + {x^3}{y^3}\,\vec k\) and \(S\) is the part of \(z = 5 - {x^2} - {y^2}\) above the plane \(z = 1\). Assume that \(S\) is oriented upwards.
    Show Solution

    Let’s start this off with a sketch of the surface.

    This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis moves off the right and slightly downward.  This is the graph of the vaguely cup shaped object given by the elliptic paraboloid in the problem statement.  It starts at z=5, is centered on the z-axis and opens in the negative z directions.  Along the base of the hemisphere is a curve, labeled C, with arrow heads indicated that it is traced out in a counter clockwise motion if viewed from above.

    In this case the boundary curve \(C\) will be where the surface intersects the plane \(z = 1\) and so will be the curve

    \[\begin{align*}1 & = 5 - {x^2} - {y^2}\\ {x^2} + {y^2} & = 4\hspace{0.25in}{\mbox{at }}z = 1\end{align*}\]

    So, the boundary curve will be the circle of radius 2 that is in the plane \(z = 1\). The parameterization of this curve is,

    \[\vec r\left( t \right) = 2\cos t\,\vec i + 2\sin t\,\vec j + \vec k,\,\,\,\,0 \le t \le 2\pi \]

    The first two components give the circle and the third component makes sure that it is in the plane \(z = 1\).

    Using Stokes’ Theorem we can write the surface integral as the following line integral.

    \[\iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\,\centerdot \,d\vec S}} = \int\limits_{C}{{\vec F\,\centerdot \,d\,\vec r}} = \int_{{\,0}}^{{\,2\pi }}{{\vec F\left( {\vec r\left( t \right)} \right)\,\centerdot \,\vec r'\left( t \right)\,dt}}\]

    So, it looks like we need a couple of quantities before we do this integral. Let’s first get the vector field evaluated on the curve. Remember that this is simply plugging the components of the parameterization into the vector field.

    \[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right) & = {\left( 1 \right)^2}\,\vec i - 3\left( {2\cos t} \right)\left( {2\sin t} \right)\,\vec j + {\left( {2\cos t} \right)^3}{\left( {2\sin t} \right)^3}\,\vec k\\ & = \vec i - 12\cos t\sin t\,\vec j + 64{\cos ^3}t{\sin ^3}t\,\vec k\end{align*}\]

    Next, we need the derivative of the parameterization and the dot product of this and the vector field.

    \[\begin{align*}\vec r'\left( t \right) & = - 2\sin t\,\vec i + 2\cos t\,\vec j\\ \vec F\left( {\vec r\left( t \right)} \right)\,\centerdot \,\vec r'\left( t \right)\, & = - 2\sin t - 24\sin t{\cos ^2}t\end{align*}\]

    We can now do the integral.

    \[\begin{align*}\iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\,\centerdot \,d\vec S}} & = \int_{{\,0}}^{{\,2\pi }}{{ - 2\sin t - 24\sin t{{\cos }^2}t\,dt}}\\ & = \left. {\left( {2\cos t + 8{{\cos }^3}t} \right)} \right|_0^{2\pi }\\ & = 0\end{align*}\]
    Example 2 Use Stokes’ Theorem to evaluate \(\displaystyle \int\limits_{C}{{\vec F\,\centerdot \,d\,\vec r}}\) where \(\vec F = {z^2}\,\vec i + {y^2}\,\vec j + x\,\vec k\) and \(C\) is the triangle with vertices \(\left( {1,0,0} \right)\), \(\left( {0,1,0} \right)\) and \(\left( {0,0,1} \right)\) with counter-clockwise rotation.
    Show Solution

    We are going to need the curl of the vector field eventually so let’s get that out of the way first.

    \[{\mathop{\rm curl}\nolimits} \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\{{z^2}}&{{y^2}}&x\end{array}} \right| = 2z\,\vec j - \vec j = \left( {2z - 1} \right)\vec j\]

    Now, all we have is the boundary curve for the surface that we’ll need to use in the surface integral. However, as noted above all we need is any surface that has this as its boundary curve. So, let’s use the following plane with upwards orientation for the surface.

    This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis moves off the right and slightly downward.  This is the graph of the triangle with vertices (0,0,1), (1,0,0) and (0,1,0).  Along the edges of it is a curve, labeled C, with arrow heads on it that indicate that it is traced out in the counter clockwise direction if we are in front of the plane and looking in towards the origin.

    Since the plane is oriented upwards this induces the positive direction on \(C\) as shown. The equation of this plane is,

    \[x + y + z = 1\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,z = g\left( {x,y} \right) = 1 - x - y\]

    Now, let’s use Stokes’ Theorem and get the surface integral set up.

    \[\begin{align*}\int\limits_{C}{{\vec F\,\centerdot \,d\,\vec r}} & = \iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\,\centerdot \,d\vec S}}\\ & = \iint\limits_{S}{{\left( {2z - 1} \right)\vec j\centerdot d\vec S}}\\ & = \iint\limits_{D}{{\left( {2z - 1} \right)\vec j\centerdot \frac{{\nabla f}}{{\left\| {\nabla f} \right\|}}\,\,\left\| {\nabla f} \right\|\,dA}}\end{align*}\]

    Okay, we now need to find a couple of quantities. First let’s get the gradient. Recall that this comes from the function of the surface.

    \[\begin{align*}f\left( {x,y,z} \right) &= z - g\left( {x,y} \right) = z - 1 + x + y\\ & \nabla f = \vec i + \vec j + \vec k\end{align*}\]

    Note as well that this also points upwards and so we have the correct direction.

    Now, \(D\) is the region in the \(xy\)-plane shown below,

    This is the 2D graph on a xy axis system of a triangle with vertices (0,0), (1,0) and (0,1).  The top left edge of the triangle is given by y=-x+1.  The bottom of the triangle is the x-axis and the left edge is the y-axis.  The triangle has been shaded in.

    We get the equation of the line by plugging in \(z = 0\) into the equation of the plane. So based on this the ranges that define \(D\) are,

    \[0 \le x \le 1\hspace{0.25in}0 \le y \le - x + 1\]

    The integral is then,

    \[\begin{align*}\int\limits_{C}{{\vec F\,\centerdot \,d\,\vec r}} & = \iint\limits_{D}{{\left( {2z - 1} \right)\vec j\centerdot \left( {\vec i + \vec j + \vec k} \right)\,dA}}\\ & = \int_{{\,0}}^{{\,1}}{{\int_{{\,0}}^{{ - x + 1}}{{2\left( {1 - x - y} \right) - 1\,dy}}\,dx}}\end{align*}\]

    Don’t forget to plug in for \(z\) since we are doing the surface integral on the plane. Finishing this out gives,

    \[\begin{align*}\int\limits_{C}{{\vec F\,\centerdot \,d\,\vec r}} & = \int_{{\,0}}^{{\,1}}{{\int_{{\,0}}^{{ - x + 1}}{{1 - 2x - 2y\,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {y - 2xy - {y^2}} \right)} \right|_0^{ - x + 1}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{{x^2} - x\,dx}}\\ & = \left. {\left( {\frac{1}{3}{x^3} - \frac{1}{2}{x^2}} \right)} \right|_0^1\\ & = - \frac{1}{6}\end{align*}\]

    In both of these examples we were able to take an integral that would have been somewhat unpleasant to deal with and by the use of Stokes’ Theorem we were able to convert it into an integral that wasn’t too bad.

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