Surface Integrals

So far, we've seen integrals over lines and paths, integrals over regions, and integrals over volumes. Now, we'll do integrals over surfaces in R3\mathbb{R}^3.

Parametric Surfaces

First, we need to discuss how to describe a surface parametrically. Recall that curves in space can be parametrized in terms of one parameter tt; surfaces in space will require two parameters uu and vv:

r⃗(u,v)=⟨x(u,v),y(u,v),z(u,v)⟩\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle

Express the following surface of a cylinder parametrically:

S={(x,y,z) : x=acos⁡θ,y=asin⁡θ,0≤θ≤2π,0≤z≤h}S = {(x,y,z)\ :\ x=a\cos\theta, y=a\sin\theta, 0 \leq \theta \leq 2\pi, 0 \leq z \leq h}

Solution

Let u=θu=θ and v=zv=z:

r⃗(u,v)=⟨acos⁡u,asin⁡u,v⟩0≤u≤2π, 0≤v≤h\ans{\begin{aligned} \vec{r}(u,v) &= \langle a \cos u, a \sin u, v \rangle\ \ &0 \leq u \leq 2\pi,\ 0 \leq v \leq h \end{aligned}}

Express the following plane parametrically: 3x−2y+z=23x-2y+z=2

Solution

Let u=xu=x and v=yv=y: r⃗(u,v)=⟨u,v,2−3u+2v⟩\ans{\vec{r}(u,v) = \langle u,v,2-3u+2v \rangle}

Express the following surface of a paraboloid parametrically: z=x2+y2    0≤z≤9z=x^2+y^2\ \ \ \ 0 \leq z \leq 9

Option A

Let u=xu=x and v=yv=y: r⃗(u,v)=⟨u,v,u2+v2⟩0≤v≤9\ans{\begin{aligned} \vec{r}(u,v) &= \langle u,v,u^2+v^2 \rangle\ \ &0 \leq v \leq 9 \end{aligned}}

Option B

Let u=θu=θ and v=r=zv=r=\sqrt{z}: r⃗(u,v)=⟨vcos⁡u,vsin⁡u,v2⟩0≤u≤2π, 0≤v≤3\ans{\begin{aligned} \vec{r}(u,v) &= \langle v \cos u, v \sin u, v^2 \rangle\ \ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 3 \end{aligned}}

Surface Integrals

Surface with vectors r_u and r_v

If a surface is expressed parametrically as r⃗(u,v)=⟨x(u,v),y(u,v),z(u,v)⟩\vec{r}(u,v)=⟨x(u,v),y(u,v),z(u,v)⟩, then

r⃗u=∂r⃗∂u=⟨∂x∂u,∂y∂u,∂z∂u⟩r⃗v=∂r⃗∂v=⟨∂x∂v,∂y∂v,∂z∂v⟩\begin{aligned} \vec{r}_u &= \dfrac{\partial \vec{r}}{\partial u} = \left\langle \dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial z}{\partial u} \right\rangle\ \vec{r}_v &= \dfrac{\partial \vec{r}}{\partial v} = \left\langle \dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v},\dfrac{\partial z}{\partial v} \right\rangle\ \end{aligned}

and the area of a differential rectangle on the surface is given by ∣r⃗u×r⃗v∣ du dv|r⃗_u×r⃗_v|\ du\ dv.

Definition

Let ff be a continuous scalar-valued function on a smooth surface SS given parametrically by r⃗(u,v)=⟨x(u,v),y(u,v),z(u,v)⟩,\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle, where a≤u≤ba≤u≤b and c≤v≤dc≤v≤d. If r⃗ur⃗_u and r⃗vr⃗_v are continuous on this region RR R={(u,v) : a≤u≤b,c≤v≤d}R = {(u,v)\ :\ a \leq u \leq b, c \leq v \leq d} then the surface integral of ff over SS is ∬Sf(x,y,z) dS=∬Rf(x(u,v),y(u,v),z(u,v)) ∣r⃗u×r⃗v∣ dA.\iint_S f(x,y,z) \ dS = \iint_R f\bigg( x(u,v),y(u,v),z(u,v) \bigg) \ |\vec{r}_u \times \vec{r}_v| \ dA.

Note

If f(x,y,z)=1f(x,y,z)=1, this integral gives the surface area of SS: Surface area =∬R∣r⃗u×r⃗v∣ dA\textrm{Surface area } = \iint_R |\vec{r}_u \times \vec{r}_v| \ dA

Examples

Find the surface area of the cylinder with radius 33 and height 55.

r⃗(u,v)=⟨3cos⁡u,3sin⁡u,v⟩0≤u≤2π, 0≤v≤5\begin{aligned} \vec{r}(u,v) &= \langle 3 \cos u, 3 \sin u, v \rangle\ \ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 5 \end{aligned}

Solution

r⃗u=⟨−3sin⁡u,3cos⁡u,0⟩r⃗v=⟨0,0,1⟩r⃗u×r⃗v=∣ı^ȷ^k^−3sin⁡u3cos⁡u0001∣=⟨3cos⁡u,3sin⁡u,0⟩∣r⃗u×r⃗v∣=3\begin{aligned} \vec{r}_u &= \langle -3 \sin u, 3 \cos u, 0 \rangle\ \vec{r}_v &= \langle 0,0,1 \rangle\ \ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\ -3 \sin u & 3 \cos u & 0\ 0 & 0 & 1 \end{vmatrix}\ \ &= \langle 3 \cos u, 3 \sin u, 0 \rangle\ \ |\vec{r}_u \times \vec{r}_v| &= 3 \end{aligned}

Therefore, the surface area is SA=∫02π∫053 dv du=∫02π15 du=30πSA = \int_0^{2\pi} \int_0^5 3 \ dv\ du = \int_0^{2\pi} 15 \ du = \ans{30\pi}

  1. Find the surface area of the plane 3x−2y+z=23x−2y+z=2 over the region 0≤x≤20≤x≤2, 0≤y≤10≤y≤1.

    2. ∫02∫0114 dv du=214\displaystyle\int_0^2 \displaystyle\int_0^1 \sqrt{14}\ dv\ du = 2\sqrt{14}

Find the surface area of the following part of a paraboloid: z=x2+y2,    0≤z≤9z=x^2+y^2, \ \ \ \ 0 \leq z \leq 9

Solution

Recall: r⃗(u,v)=⟨vcos⁡u,vsin⁡u,v2⟩0≤u≤2π, 0≤v≤3\begin{aligned} \vec{r}(u,v) &= \langle v \cos u, v \sin u, v^2 \rangle\ \ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 3 \end{aligned}

Then r⃗u=⟨−vsin⁡u,vcos⁡u,0⟩r⃗v=⟨cos⁡u,sin⁡u,2v⟩r⃗u×r⃗v=∣ı^ȷ^k^−vsin⁡u−vcos⁡u0cos⁡usin⁡u2v∣=⟨2v2cos⁡u,2v2sin⁡u,−v⟩∣r⃗u×r⃗v∣=v4v2+1\begin{aligned} \vec{r}_u &= \langle -v \sin u, v \cos u, 0 \rangle\ \vec{r}_v &= \langle \cos u, \sin u, 2v \rangle\ \ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\ -v \sin u & -v \cos u & 0\ \cos u & \sin u & 2v \end{vmatrix}\ \ &= \langle 2v^2 \cos u, 2v^2 \sin u, -v \rangle\ \ |\vec{r}_u \times \vec{r}_v| &= v\sqrt{4v^2+1} \end{aligned}

Therefore, the surface area is SA=∫03∫02πv4v2+1 du dv=2π∫03v4v2+1 dv=π6[(4v2+1)3/2]03=π6(373/2−1)\begin{aligned} SA &= \int_0^3 \int_0^{2\pi} v\sqrt{4v^2+1} \ du\ dv\ &= 2\pi \int_0^3 v\sqrt{4v^2+1} \ dv\ &= \dfrac{\pi}{6} \bigg[(4v^2+1)^{3/2}\bigg]_0^3\ &= \ans{\dfrac{\pi}{6}\left(37^{3/2}-1\right)} \end{aligned}

Evaluate the integral ∬Syz dS\displaystyle\iint_S \dfrac{y}{z} \ dS where SS is the following parametric surface: r⃗(u,v)=⟨u2,usin⁡v,ucos⁡v⟩,    0≤u≤1,  0≤v≤π3\vec{r}(u,v) = \langle u^2, u \sin v, u \cos v \rangle, \ \ \ \ 0 \leq u \leq 1, \ \ 0 \leq v \leq \dfrac{\pi}{3}

Solution

First, note that y=usin⁡vy=u \sin v and z=ucos⁡vz=u \cos v, so

r⃗u=⟨2u,sin⁡v,cos⁡v⟩r⃗v=⟨0,ucos⁡v,−usin⁡v⟩r⃗u×r⃗v=⟨−u,2u2sin⁡v,2u2cos⁡v⟩∣r⃗u×r⃗v∣=u1+4u2\begin{aligned} \vec{r}_u &= \langle 2u, \sin v, \cos v \rangle\ \vec{r}_v &= \langle 0, u \cos v, -u \sin v \rangle\ \ \vec{r}_u \times \vec{r}_v &= \langle -u, 2u^2 \sin v, 2u^2 \cos v \rangle\ \ |\vec{r}_u \times \vec{r}_v| &= u\sqrt{1+4u^2} \end{aligned}

Therefore, this surface integral is

∬Syz dS=∫0π/3∫01sin⁡vcos⁡v⋅u1+4u2 du dv=∫0π/3[112(4u2+1)3/2∣01] sin⁡vcos⁡v dv=112(53/2−1)∫0π/3sin⁡vcos⁡v dv=112(53/2−1)[−ln⁡∣cos⁡v∣]0π/3=−112(53/2−1)ln⁡12\begin{aligned} \iint_S \dfrac{y}{z} \ dS &= \int_0^{\pi/3} \int_0^1 \dfrac{\sin v}{\cos v} \cdot u \sqrt{1+4u^2}\ du\ dv\ &= \int_0^{\pi/3} \left[\dfrac{1}{12} (4u^2+1)^{3/2} \bigg|_0^1 \right] \ \dfrac{\sin v}{\cos v} \ dv\ &= \dfrac{1}{12} (5^{3/2}-1) \int_0^{\pi/3} \dfrac{\sin v}{\cos v} \ dv\ &= \dfrac{1}{12} (5^{3/2}-1) \bigg[-\ln |\cos v|\bigg]_0^{\pi/3}\ &= \ans{-\dfrac{1}{12}(5^{3/2}-1)\ln \dfrac{1}{2}} \end{aligned}

Explicitly Defined Surfaces

If a surface is defined explicitly as z=g(x,y)z=g(x,y), we can parameterize it using u=xu=x and v=yv=y: r⃗(u,v)=⟨u,v,g(u,v)⟩\vec{r} (u,v) = \langle u,v,g(u,v) \rangle

Then r⃗u=⟨1,0,zx⟩r⃗_u=⟨1,0,z_x⟩ and r⃗v=⟨0,1,zy⟩r⃗_v=⟨0,1,z_y⟩: r⃗u×r⃗v=∣ı^ȷ^k^10zx01zy∣=⟨−zx,−zy,1⟩  ⟹  ∣r⃗u×r⃗v∣=zx2+zy2+1\begin{aligned} \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\ 1 & 0 & z_x\ 0 & 1 & z_y \end{vmatrix}\ \ &= \langle -z_x,-z_y,1 \rangle\ \ &\implies |\vec{r}_u \times \vec{r}_v| = \sqrt{z_x^2+z_y^2+1} \end{aligned}

The surface integral can thus be written as ∬Sf(x,y,z) dS=∬Rf(x,y,g(x,y))zx2+zy2+1 dA\iint_S f(x,y,z) \ dS = \iint_R f\bigg(x,y,g(x,y)\bigg) \sqrt{z_x^2+z_y^2+1} \ dA

Again, if f(x,y,z)=1f(x,y,z)=1, this integral gives the surface area.

Evaluate the integral ∬Sxy dS\displaystyle\iint_S xy \ dS where SS is the triangular surface with vertices (1,0,0)(1,0,0), (0,2,0)(0,2,0), and (0,0,2)(0,0,2).

Triangular surface

Solution

First, find the equation of this plane by taking the cross product of two vectors in the plane:

PQ⃗=⟨−1,0,2⟩PR⃗=⟨−1,2,0⟩PQ⃗×PR⃗=⟨−4,−2,−2⟩ or ⟨2,1,1⟩\begin{aligned} \vec{PQ} &= \langle -1,0,2 \rangle\ \vec{PR} &= \langle -1,2,0 \rangle\ \ \vec{PQ} \times \vec{PR} &= \langle -4,-2,-2 \rangle \textrm{ or } \langle 2,1,1 \rangle \end{aligned}

Therefore, the equation of this plane is 2(x−1)+1(y−0)+1(z−0)2(x−1)+1(y−0)+1(z−0) or z=−2x−y+2z=−2x−y+2, so zx=−2z_x=−2 and zy=−1z_y=−1. ∬Sxy dS=∫01∫0−2x+2xy(−2)2+(−1)2+1 dy dx≈0.408\iint_S xy\ dS = \int_0^1 \int_0^{-2x+2} xy \sqrt{(-2)^2+(-1)^2+1} \ dy\ dx \approx \ans{0.408}

Surface Integrals in Vector Fields: Flux

There is a special surface integral known as the flux, which measures how much of a vector field F⃗\vec{F} passes through a surface. Physical applications of this include fluid flow, heat flow, electrical current flow, etc. (you may run across heat flux or current flux at some point in the future).

Flux =∬SF⃗⋅n^ dS=∬RF⃗⋅(r⃗u×r⃗v) dA\textrm{Flux } = \ans{\iint_S \vec{F} \cdot \hat{n} \ dS = \iint_R \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \ dA}

because n^=r⃗u×r⃗v∣r⃗u×r⃗v∣.\hat{n} = \dfrac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|}.

If the surface is defined explicitly by z=g(x,y)z=g(x,y), then ∬SF⃗⋅n^ dS=∬R−fzx−gzy+h dA\ans{\iint_S \vec{F} \cdot \hat{n} \ dS = \iint_R -f z_x - g z_y + h \ dA} where F⃗=⟨f,g,h⟩.\vec{F}=⟨f,g,h⟩.

Rainfall on a Sloped Roof

If F⃗=⟨0,0,−1⟩\vec{F}=⟨0,0,−1⟩ and z=4−2x−yz=4−2x−y for 0≤x≤20≤x≤2, 0≤y≤4−2x0≤y≤4−2x, find the flux of F⃗\vec{F} through this surface.

Solution

zx=−2zy=−1∬SF⃗⋅n^ dS=∫02∫04−2x−1 dy dx=∫022x−4 dx=x2−4x∣02=−4\begin{aligned} z_x &= -2\ z_y &= -1\ \ \iint_S \vec{F} \cdot \hat{n}\ dS &= \int_0^2 \int_0^{4-2x} -1 \ dy\ dx\ &= \int_0^2 2x-4 \ dx\ &= x^2-4x \bigg|_0^2 = \ans{-4} \end{aligned}

This answer gives the amount of rain that passes through this surface in one unit of time (note that it is negative, since the rain is moving downward).