How to Calculate Surface Integrals

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. 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. The derivation works in any coordinate system. See the tips for the derivation in cylindrical coordinates.

Visualize a surface integral. The surface consists of infinitesimal patches that are approximately flat. As you can see, the way we integrate over a domain works the same way, and the fact that a surface element denotes orientation as well reflects that surface integrals are a powerful generalization of area integrals.

Surface Area

Calculate the surface area of the function z = 4 − x 2 − y 2 {\displaystyle z=4-x^{2}-y^{2}} z=4-x^{{2}}-y^{{2}} above the xy-plane. Finding the surface area involves finding the integral below. We only care about the area of the surface, not its orientation, so we find its magnitude. ∫ S | d S | = ∫ A 1 + ( ∂ z ∂ x ) 2 + ( ∂ z ∂ y ) 2 d A {\displaystyle \int _{S}|\mathrm {d} \mathbf {S} |=\int _{A}{\sqrt {1+\left({\frac {\partial z}{\partial x}}\right)^{2}+\left({\frac {\partial z}{\partial y}}\right)^{2}}}\,\mathrm {d} A} \int _{{S}}|{\mathrm {d}}{\mathbf {S}}|=\int _{{A}}{\sqrt {1+\left({\frac {\partial z}{\partial x}}\right)^{{2}}+\left({\frac {\partial z}{\partial y}}\right)^{{2}}}}\,{\mathrm {d}}A

Find the magnitude of the surface element. Recall from part 1 that 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, where z = f ( x , y ) . {\displaystyle z=f(x,y).} z=f(x,y). d S = 2 x i + 2 y j + k {\displaystyle \mathrm {d} \mathbf {S} =2x\mathbf {i} +2y\mathbf {j} +\mathbf {k} } {\mathrm {d}}{\mathbf {S}}=2x{\mathbf {i}}+2y{\mathbf {j}}+{\mathbf {k}} | d S | = 4 x 2 + 4 y 2 + 1 d A {\displaystyle |\mathrm {d} \mathbf {S} |={\sqrt {4x^{2}+4y^{2}+1}}\mathrm {d} A} |{\mathrm {d}}{\mathbf {S}}|={\sqrt {4x^{{2}}+4y^{{2}}+1}}{\mathrm {d}}A

Set the boundaries. The boundary on the xy-plane is a circle of radius 2. This means that we should evaluate in polar coordinates too. ∫ S | d S | = ∫ 0 2 r d r ∫ 0 2 π d θ 4 r 2 + 1 {\displaystyle \int _{S}|\mathrm {d} \mathbf {S} |=\int _{0}^{2}r\mathrm {d} r\int _{0}^{2\pi }\mathrm {d} \theta {\sqrt {4r^{2}+1}}} \int _{{S}}|{\mathrm {d}}{\mathbf {S}}|=\int _{{0}}^{{2}}r{\mathrm {d}}r\int _{{0}}^{{2\pi }}{\mathrm {d}}\theta {\sqrt {4r^{{2}}+1}}

Evaluate using any means possible. U-substitution is the way to go. ∫ S | d S | = 2 π ∫ 0 2 r 4 r 2 + 1 d r , u = 1 + 4 r 2 = π 4 ∫ 1 17 u 1 / 2 d u = π 6 ( 17 3 / 2 − 1 ) {\displaystyle {\begin{aligned}\int _{S}|\mathrm {d} \mathbf {S} |&=2\pi \int _{0}^{2}r{\sqrt {4r^{2}+1}}\mathrm {d} r,\ \ \ u=1+4r^{2}\\&={\frac {\pi }{4}}\int _{1}^{17}u^{1/2}\mathrm {d} u\\&={\frac {\pi }{6}}(17^{3/2}-1)\end{aligned}}} {\begin{aligned}\int _{{S}}|{\mathrm {d}}{\mathbf {S}}|&=2\pi \int _{{0}}^{{2}}r{\sqrt {4r^{{2}}+1}}{\mathrm {d}}r,\ \ \ u=1+4r^{{2}}\\&={\frac {\pi }{4}}\int _{{1}}^{{17}}u^{{1/2}}{\mathrm {d}}u\\&={\frac {\pi }{6}}(17^{{3/2}}-1)\end{aligned}}