How to Use Stokes' Theorem

Deriving the Surface Element

Consider an arbitrary vector function r {\displaystyle \mathbf {r} } {\mathbf {r}}. Below, we let z = f ( x , y ) . {\displaystyle z=f(x,y).} z=f(x,y). r = x i + y j + f ( x , y ) k {\displaystyle \mathbf {r} =x\mathbf {i} +y\mathbf {j} +f(x,y)\mathbf {k} } {\mathbf {r}}=x{\mathbf {i}}+y{\mathbf {j}}+f(x,y){\mathbf {k}}

Calculate differentials. For d r x , y {\displaystyle \mathrm {d} \mathbf {r} _{x},\,y} {\mathrm {d}}{\mathbf {r}}_{{x}},\,y is being held constant, and vice versa. We use the notation f x = ∂ f ( x , y ) ∂ x . {\displaystyle f_{x}={\frac {\partial f(x,y)}{\partial x}}.} f_{{x}}={\frac {\partial f(x,y)}{\partial x}}. d r x = d x i + f x d x k {\displaystyle \mathrm {d} \mathbf {r} _{x}=\mathrm {d} x\mathbf {i} +f_{x}\mathrm {d} x\mathbf {k} } {\mathrm {d}}{\mathbf {r}}_{{x}}={\mathrm {d}}x{\mathbf {i}}+f_{{x}}{\mathrm {d}}x{\mathbf {k}} d r y = d y j + f y d y k {\displaystyle \mathrm {d} \mathbf {r} _{y}=\mathrm {d} y\mathbf {j} +f_{y}\mathrm {d} y\mathbf {k} } {\mathrm {d}}{\mathbf {r}}_{{y}}={\mathrm {d}}y{\mathbf {j}}+f_{{y}}{\mathrm {d}}y{\mathbf {k}}

Take the cross product of the two differentials. Surface integrals are a generalization of line integrals. A surface element therefore contains information about both its area and orientation. Thus, the goal is to compute a cross product. d S = d r x × d r y = | i j k d x 0 f x d x 0 d y f y d y | = − f x d x d y i − f y d x d y j + d x d y k {\displaystyle {\begin{aligned}\mathrm {d} \mathbf {S} &=\mathrm {d} \mathbf {r} _{x}\times \mathrm {d} \mathbf {r} _{y}\\&={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\\mathrm {d} x&0&f_{x}\mathrm {d} x\\0&\mathrm {d} y&f_{y}\mathrm {d} y\end{vmatrix}}\\&=-f_{x}\mathrm {d} x\mathrm {d} y\mathbf {i} -f_{y}\mathrm {d} x\mathrm {d} y\mathbf {j} +\mathrm {d} x\mathrm {d} y\mathbf {k} \end{aligned}}} {\begin{aligned}{\mathrm {d}}{\mathbf {S}}&={\mathrm {d}}{\mathbf {r}}_{{x}}\times {\mathrm {d}}{\mathbf {r}}_{{y}}\\&={\begin{vmatrix}{\mathbf {i}}&{\mathbf {j}}&{\mathbf {k}}\\{\mathrm {d}}x&0&f_{{x}}{\mathrm {d}}x\\0&{\mathrm {d}}y&f_{{y}}{\mathrm {d}}y\end{vmatrix}}\\&=-f_{{x}}{\mathrm {d}}x{\mathrm {d}}y{\mathbf {i}}-f_{{y}}{\mathrm {d}}x{\mathrm {d}}y{\mathbf {j}}+{\mathrm {d}}x{\mathrm {d}}y{\mathbf {k}}\end{aligned}} d S = ( − f x i − f y j + k ) d x d y {\displaystyle \mathrm {d} \mathbf {S} =(-f_{x}\mathbf {i} -f_{y}\mathbf {j} +\mathbf {k} )\mathrm {d} x\mathrm {d} y} {\mathrm {d}}{\mathbf {S}}=(-f_{{x}}{\mathbf {i}}-f_{{y}}{\mathbf {j}}+{\mathbf {k}}){\mathrm {d}}x{\mathrm {d}}y The formula above is the surface element for general surfaces defined by z = f ( x , y ) . {\displaystyle z=f(x,y).} z=f(x,y). It is important to note that the nature of surfaces (more accurately, the cross product) still allows one ambiguity - the way the normal vector is pointing. The result that we have derived applies to outward normals, as recognized by the positive k {\displaystyle \mathbf {k} } {\mathbf {k}} component, and for most applications, this will always be the case.

Example 1

Find the surface integral of G {\displaystyle \mathbf {G} } {\mathbf {G}} over the surface z = 12 − 4 x 2 − 3 y 2 {\displaystyle z=12-4x^{2}-3y^{2}} z=12-4x^{{2}}-3y^{{2}}. The surface below has a boundary of an ellipse, not a circle. If we choose to do the surface integral, then we will need to use Jacobian change of variables in order to properly convert into polar coordinates. Therefore, we will choose to parameterize the boundary directly. G = ( 2 x 2 y − 4 x z ) i + y j + x 3 z k {\displaystyle \mathbf {G} =(2x^{2}y-4xz)\mathbf {i} +y\mathbf {j} +x^{3}z\mathbf {k} } {\mathbf {G}}=(2x^{{2}}y-4xz){\mathbf {i}}+y{\mathbf {j}}+x^{{3}}z{\mathbf {k}}

Parameterize the boundary. As always, verify that the chosen parameters work before proceeding. x = 3 cos ⁡ t {\displaystyle x={\sqrt {3}}\cos t} x={\sqrt {3}}\cos t y = 2 sin ⁡ t {\displaystyle y=2\sin t} y=2\sin t

Calculate differentials. d x = − 3 sin ⁡ t {\displaystyle \mathrm {d} x=-{\sqrt {3}}\sin t} {\mathrm {d}}x=-{\sqrt {3}}\sin t d y = 2 cos ⁡ t {\displaystyle \mathrm {d} y=2\cos t} {\mathrm {d}}y=2\cos t

Substitute these parameters into the vector field, and take the resulting dot product G ⋅ d r {\displaystyle \mathbf {G} \cdot \mathrm {d} \mathbf {r} } {\mathbf {G}}\cdot {\mathrm {d}}{\mathbf {r}}. Since our boundary is on the xy-plane, z = 0 , {\displaystyle z=0,} z=0, so cross out all terms that contain z . {\displaystyle z.} z. Additionally, we are performing a closed loop integral, so our interval is [ 0 , 2 π ] . {\displaystyle [0,2\pi ].} [0,2\pi ]. ∮ G ⋅ d r = ∫ 0 2 π ( 2 ⋅ 3 cos 2 ⁡ t ⋅ 2 sin ⁡ t ⋅ − 3 sin ⁡ t + 4 sin ⁡ t cos ⁡ t ) d t = ∫ 0 2 π ( − 12 3 cos 2 ⁡ t sin 2 ⁡ t + 4 sin ⁡ t cos ⁡ t ) d t {\displaystyle {\begin{aligned}\oint \mathbf {G} \cdot \mathrm {d} \mathbf {r} &=\int _{0}^{2\pi }(2\cdot 3\cos ^{2}t\cdot 2\sin t\cdot -{\sqrt {3}}\sin t+4\sin t\cos t)\mathrm {d} t\\&=\int _{0}^{2\pi }(-12{\sqrt {3}}\cos ^{2}t\sin ^{2}t+4\sin t\cos t)\mathrm {d} t\end{aligned}}} {\begin{aligned}\oint {\mathbf {G}}\cdot {\mathrm {d}}{\mathbf {r}}&=\int _{{0}}^{{2\pi }}(2\cdot 3\cos ^{{2}}t\cdot 2\sin t\cdot -{\sqrt {3}}\sin t+4\sin t\cos t){\mathrm {d}}t\\&=\int _{{0}}^{{2\pi }}(-12{\sqrt {3}}\cos ^{{2}}t\sin ^{{2}}t+4\sin t\cos t){\mathrm {d}}t\end{aligned}}

Cancel out terms. The second term is 0 if we perform a u-substitution. ∫ 0 2 π ( − 12 3 cos 2 ⁡ t sin 2 ⁡ t + 4 sin ⁡ t cos ⁡ t ) d t {\displaystyle \int _{0}^{2\pi }(-12{\sqrt {3}}\cos ^{2}t\sin ^{2}t+{\cancel {4\sin t\cos t}})\mathrm {d} t} \int _{{0}}^{{2\pi }}(-12{\sqrt {3}}\cos ^{{2}}t\sin ^{{2}}t+{\cancel {4\sin t\cos t}}){\mathrm {d}}t

Evaluate using any means possible. It is useful to memorize ∫ 0 2 π cos 2 ⁡ t sin 2 ⁡ t d t = π 4 . {\displaystyle \int _{0}^{2\pi }\cos ^{2}t\sin ^{2}t\mathrm {d} t={\frac {\pi }{4}}.} \int _{{0}}^{{2\pi }}\cos ^{{2}}t\sin ^{{2}}t{\mathrm {d}}t={\frac {\pi }{4}}. − 12 3 ∫ 0 2 π cos 2 ⁡ t sin 2 ⁡ t d t = − 3 3 π {\displaystyle -12{\sqrt {3}}\int _{0}^{2\pi }\cos ^{2}t\sin ^{2}t\mathrm {d} t=-3{\sqrt {3}}\pi } -12{\sqrt {3}}\int _{{0}}^{{2\pi }}\cos ^{{2}}t\sin ^{{2}}t{\mathrm {d}}t=-3{\sqrt {3}}\pi To check that this answer is correct, simply do the surface integral. The process will be longer, since you have to take the curl of a vector field and do Jacobians when you convert to the area integral.

Example 2

Verify Stokes' theorem. Use the surface z = 6 − x 2 − y 2 {\displaystyle z=6-x^{2}-y^{2}} z=6-x^{{2}}-y^{{2}} above the xy-plane with the given vector field below. F = ( x 2 − 2 y ) i + ( 3 x − 2 x z ) j + ( x 2 y 2 − 4 y z ) k {\displaystyle \mathbf {F} =(x^{2}-2y)\mathbf {i} +(3x-2xz)\mathbf {j} +(x^{2}y^{2}-4yz)\mathbf {k} } {\mathbf {F}}=(x^{{2}}-2y){\mathbf {i}}+(3x-2xz){\mathbf {j}}+(x^{{2}}y^{{2}}-4yz){\mathbf {k}} The goal of verification is to evaluate both integrals and check that their answers are the same. First, we will parameterize the boundary and compute the line integral. Then, we'll evaluate the surface integral. With enough practice using Stokes' theorem, you will be able to rewrite a problem into something that is easier to solve.

Parameterize the boundary. When we set z = 0 , {\displaystyle z=0,} z=0, we find that the boundary is a circle of radius 6 {\displaystyle {\sqrt {6}}} {\sqrt {6}} on the xy-plane. Therefore, the following parameters are appropriate. These are the components of r . {\displaystyle \mathbf {r} .} {\mathbf {r}}. x = 6 cos ⁡ t {\displaystyle x={\sqrt {6}}\cos t} x={\sqrt {6}}\cos t y = 6 sin ⁡ t {\displaystyle y={\sqrt {6}}\sin t} y={\sqrt {6}}\sin t

Calculate differentials. d x = − 6 sin ⁡ t d t {\displaystyle \mathrm {d} x=-{\sqrt {6}}\sin t\mathrm {d} t} {\mathrm {d}}x=-{\sqrt {6}}\sin t{\mathrm {d}}t d y = 6 cos ⁡ t d t {\displaystyle \mathrm {d} y={\sqrt {6}}\cos t\mathrm {d} t} {\mathrm {d}}y={\sqrt {6}}\cos t{\mathrm {d}}t

Calculate the dot product F ⋅ d r {\displaystyle \mathbf {F} \cdot \mathrm {d} \mathbf {r} } {\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {r}}. The vector field contains terms with z {\displaystyle z} z in them, but since on the xy-plane, z = 0 , {\displaystyle z=0,} z=0, neglect those terms. F ⋅ d r = ( 6 cos 2 ⁡ t − 2 6 sin ⁡ t ) ( − 6 sin ⁡ t d t ) + ( 3 6 cos ⁡ t ) ( 6 cos ⁡ t d t ) = ( − 6 6 cos 2 ⁡ t sin ⁡ t + 12 sin 2 ⁡ t + 18 cos 2 ⁡ t ) d t {\displaystyle {\begin{aligned}\mathbf {F} \cdot \mathrm {d} \mathbf {r} &=(6\cos ^{2}t-2{\sqrt {6}}\sin t)(-{\sqrt {6}}\sin t\mathrm {d} t)+(3{\sqrt {6}}\cos t)({\sqrt {6}}\cos t\mathrm {d} t)\\&=(-6{\sqrt {6}}\cos ^{2}t\sin t+12\sin ^{2}t+18\cos ^{2}t)\mathrm {d} t\end{aligned}}} {\begin{aligned}{\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {r}}&=(6\cos ^{{2}}t-2{\sqrt {6}}\sin t)(-{\sqrt {6}}\sin t{\mathrm {d}}t)+(3{\sqrt {6}}\cos t)({\sqrt {6}}\cos t{\mathrm {d}}t)\\&=(-6{\sqrt {6}}\cos ^{{2}}t\sin t+12\sin ^{{2}}t+18\cos ^{{2}}t){\mathrm {d}}t\end{aligned}}

Set the boundaries and simplify the integrand. Stokes' theorem tells us that t {\displaystyle t} t is being integrated on the interval [ 0 , 2 π ] . {\displaystyle [0,2\pi ].} [0,2\pi ]. It is useful to recognize that ∫ 0 2 π sin ⁡ t d t = 0 , {\displaystyle \int _{0}^{2\pi }\sin t\mathrm {d} t=0,} \int _{{0}}^{{2\pi }}\sin t{\mathrm {d}}t=0, which allows us to annihilate that term. Even though it is being multiplied by cos 2 ⁡ t , {\displaystyle \cos ^{2}t,} \cos ^{{2}}t, that does not affect sin ⁡ t {\displaystyle \sin t} \sin t being odd over the interval [ 0 , 2 π ] {\displaystyle [0,2\pi ]} [0,2\pi ] because cos 2 ⁡ t {\displaystyle \cos ^{2}t} \cos ^{{2}}t is even. ∫ ∂ S F ⋅ d r = ∫ 0 2 π ( 12 sin 2 ⁡ t + 18 cos 2 ⁡ t ) d t {\displaystyle \int _{\partial S}\mathbf {F} \cdot \mathrm {d} \mathbf {r} =\int _{0}^{2\pi }(12\sin ^{2}t+18\cos ^{2}t)\mathrm {d} t} \int _{{\partial S}}{\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {r}}=\int _{{0}}^{{2\pi }}(12\sin ^{{2}}t+18\cos ^{{2}}t){\mathrm {d}}t

Evaluate using any means possible. Here, we recognize that ∫ 0 2 π sin 2 ⁡ t d t = ∫ 0 2 π cos 2 ⁡ t d t = π , {\displaystyle \int _{0}^{2\pi }\sin ^{2}t\mathrm {d} t=\int _{0}^{2\pi }\cos ^{2}t\mathrm {d} t=\pi ,} \int _{{0}}^{{2\pi }}\sin ^{{2}}t{\mathrm {d}}t=\int _{{0}}^{{2\pi }}\cos ^{{2}}t{\mathrm {d}}t=\pi , which, while they can be found using trig identities, are worth memorizing regardless. ∫ 0 2 π ( 12 sin 2 ⁡ t + 18 cos 2 ⁡ t ) d t = 30 π {\displaystyle \int _{0}^{2\pi }(12\sin ^{2}t+18\cos ^{2}t)\mathrm {d} t=30\pi } \int _{{0}}^{{2\pi }}(12\sin ^{{2}}t+18\cos ^{{2}}t){\mathrm {d}}t=30\pi

Find the surface element d S {\displaystyle \mathrm {d} \mathbf {S} } {\mathrm {d}}{\mathbf {S}}. We recall the formula converting the surface integral into an easier-to-manage area integral as d S = ( − f x i − f y j + k ) d A . {\displaystyle \mathrm {d} \mathbf {S} =(-f_{x}\mathbf {i} -f_{y}\mathbf {j} +\mathbf {k} )\mathrm {d} A.} {\mathrm {d}}{\mathbf {S}}=(-f_{{x}}{\mathbf {i}}-f_{{y}}{\mathbf {j}}+{\mathbf {k}}){\mathrm {d}}A. In this case, f {\displaystyle f} f refers to the surface z = f ( x , y ) = 6 − x 2 − y 2 . {\displaystyle z=f(x,y)=6-x^{2}-y^{2}.} z=f(x,y)=6-x^{{2}}-y^{{2}}. d S = ( 2 x i + 2 y j + k ) d A {\displaystyle \mathrm {d} \mathbf {S} =(2x\mathbf {i} +2y\mathbf {j} +\mathbf {k} )\mathrm {d} A} {\mathrm {d}}{\mathbf {S}}=(2x{\mathbf {i}}+2y{\mathbf {j}}+{\mathbf {k}}){\mathrm {d}}A

Find the curl of F , {\displaystyle \mathbf {F} ,} {\mathbf {F}}, and compute the resulting dot product ( ∇ × F ) ⋅ d S {\displaystyle (\nabla \times \mathbf {F} )\cdot \mathrm {d} \mathbf {S} } (\nabla \times {\mathbf {F}})\cdot {\mathrm {d}}{\mathbf {S}}. During the dot product, we find that we have three variables, yet we are integrating over just two dimensions. Simply substitute z = 6 − x 2 − y 2 {\displaystyle z=6-x^{2}-y^{2}} z=6-x^{{2}}-y^{{2}} to solve this. ∇ × F = | i j k ∂ / ∂ x ∂ / ∂ y ∂ / ∂ z x 2 − 2 y 3 x − 2 x z x 2 y 2 − 4 y z | = ( 2 x 2 y − 4 z + 2 x ) i − ( 2 x y 2 ) j + ( 5 − 2 z ) k {\displaystyle {\begin{aligned}\nabla \times \mathbf {F} &={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\\partial /\partial x&\partial /\partial y&\partial /\partial z\\x^{2}-2y&3x-2xz&x^{2}y^{2}-4yz\end{vmatrix}}\\&=(2x^{2}y-4z+2x)\mathbf {i} -(2xy^{2})\mathbf {j} +(5-2z)\mathbf {k} \end{aligned}}} {\begin{aligned}\nabla \times {\mathbf {F}}&={\begin{vmatrix}{\mathbf {i}}&{\mathbf {j}}&{\mathbf {k}}\\\partial /\partial x&\partial /\partial y&\partial /\partial z\\x^{{2}}-2y&3x-2xz&x^{{2}}y^{{2}}-4yz\end{vmatrix}}\\&=(2x^{{2}}y-4z+2x){\mathbf {i}}-(2xy^{{2}}){\mathbf {j}}+(5-2z){\mathbf {k}}\end{aligned}} ∫ S ( ∇ × F ) ⋅ d S = ∫ D ( 4 x 3 y − 8 x z + 4 x 2 − 4 x y 3 + 5 − 2 z ) d A {\displaystyle \int _{S}(\nabla \times \mathbf {F} )\cdot \mathrm {d} \mathbf {S} =\int _{D}(4x^{3}y-8xz+4x^{2}-4xy^{3}+5-2z)\mathrm {d} A} \int _{{S}}(\nabla \times {\mathbf {F}})\cdot {\mathrm {d}}{\mathbf {S}}=\int _{{D}}(4x^{{3}}y-8xz+4x^{{2}}-4xy^{{3}}+5-2z){\mathrm {d}}A

Cancel out terms. The function f ( x , y ) = 6 − x 2 − y 2 {\displaystyle f(x,y)=6-x^{2}-y^{2}} f(x,y)=6-x^{{2}}-y^{{2}} is symmetric over both the x {\displaystyle x} x and y {\displaystyle y} y axes. Therefore, any terms with an odd function of either variable will cancel out. In this problem, notice that f ( x , y ) {\displaystyle f(x,y)} f(x,y) is an even function. Therefore, we don't even need to do the multiplication for the 8 x z {\displaystyle 8xz} 8xz term, because 8 x {\displaystyle 8x} 8x is odd, so the entire term cancels out. This step greatly simplifies the integral to be evaluated. ∫ D ( 4 x 3 y − 8 x z + 4 x 2 − 4 x y 3 + 5 − 12 + 2 x 2 + 2 y 2 ) d A {\displaystyle \int _{D}({\cancel {4x^{3}y}}-{\cancel {8xz}}+4x^{2}-{\cancel {4xy^{3}}}+5-12+2x^{2}+2y^{2})\mathrm {d} A} \int _{{D}}({\cancel {4x^{{3}}y}}-{\cancel {8xz}}+4x^{{2}}-{\cancel {4xy^{{3}}}}+5-12+2x^{{2}}+2y^{{2}}){\mathrm {d}}A

Simplify and convert to polar coordinates. Our problem has now been reduced to an area integral on the xy-plane, for we have taken advantage of Stokes' theorem and recognized that this "surface" - the disk on the plane - will yield the same result as our elliptic paraboloid. ∫ D ( 6 x 2 + 2 y 2 − 7 ) d A = ∫ 0 6 r d r ∫ 0 2 π d θ ( 6 r 2 cos 2 ⁡ θ + 2 r 2 sin 2 ⁡ θ − 7 ) {\displaystyle \int _{D}(6x^{2}+2y^{2}-7)\mathrm {d} A=\int _{0}^{\sqrt {6}}r\mathrm {d} r\int _{0}^{2\pi }\mathrm {d} \theta (6r^{2}\cos ^{2}\theta +2r^{2}\sin ^{2}\theta -7)} \int _{{D}}(6x^{{2}}+2y^{{2}}-7){\mathrm {d}}A=\int _{{0}}^{{{\sqrt {6}}}}r{\mathrm {d}}r\int _{{0}}^{{2\pi }}{\mathrm {d}}\theta (6r^{{2}}\cos ^{{2}}\theta +2r^{{2}}\sin ^{{2}}\theta -7)

Evaluate using any means possible. ∫ 0 6 r d r ( 8 r 2 π − 14 π ) = π ∫ 0 6 ( 8 r 3 − 14 r ) d r = π ( 2 ⋅ 36 − 7 ⋅ 6 ) = 30 π {\displaystyle {\begin{aligned}\int _{0}^{\sqrt {6}}r\mathrm {d} r(8r^{2}\pi -14\pi )&=\pi \int _{0}^{\sqrt {6}}(8r^{3}-14r)\mathrm {d} r\\&=\pi (2\cdot 36-7\cdot 6)\\&=30\pi \end{aligned}}} {\begin{aligned}\int _{{0}}^{{{\sqrt {6}}}}r{\mathrm {d}}r(8r^{{2}}\pi -14\pi )&=\pi \int _{{0}}^{{{\sqrt {6}}}}(8r^{{3}}-14r){\mathrm {d}}r\\&=\pi (2\cdot 36-7\cdot 6)\\&=30\pi \end{aligned}} Our answer agrees with our answer obtained in step 6, so Stokes' theorem has been verified.