# Surface Integrals

## Motivation

Our goal is to generalize the concept of double integral to the one in which domain of integration is a piece of a surface. When dealing with double integrals, we always had a piece of the plane as a domain. Planes are flat, and we want to deal with general surfaces now. We will always describe surfaces using their parametrizations, and you may want to review this material in the section Vector fields and parametric equations of curves and surfaces.

## Surface integrals of functions

### Integral sums

Assume that $$\sigma$$ is a surface parametric equations: $$\overrightarrow R(t,s)=\langle \phi(t,s),\psi(t,s),\theta(t,s)\rangle$$, as parameters $$t$$ and $$s$$ belong to some domain $$D$$.

Consider the partition of the domain $$D$$ in smaller domains $$D=D_1\cup D_2\cup\dots\cup D_n$$. Assume that $$P_1$$, $$\dots$$, $$P_n$$ is a sequence of points such that $$P_i\in D_i$$, and let $$A_i$$ be the area of the portion of the surface corresponding to the domain $$D_i$$. The integral sum $$S(f, D_1,\dots, D_n,P_1,\dots, P_n)$$ corresponding to the partition $$D_1$$, $$\dots$$, $$D_n$$ with the points $$P_1$$, $$\dots$$, $$P_n$$ is defined as: $S(f,D_1,\dots, D_n,P_1,\dots, P_n)=f(P_1)\cdot A_1+f(P_2)\cdot A_2+\cdots + f(P_n)\cdot A_n.$ The diameter of the partition is defined as: $\delta(D_1,\dots, D_n)=\max\{A_1,\dots, A_n\}.$

Definition: Surface integral of a function Let $$\sigma$$ be the surface with the parametrization $$x=\phi(t,s)$$, $$y=\psi(t,s)$$, $$z=\theta(t,s)$$, where $$t,s\in D$$ for some $$D\subseteq \mathbb R^2$$. The function $$f:\sigma\to\mathbb R$$ has a surface integral and its integral over the surface $$\sigma$$ is equal to $$I$$ if for each $$\varepsilon > 0$$ there exists $$\delta > 0$$ such that for every partition $$D_1,\dots, D_n$$ with $$\delta(D_1,\dots, D_{n}) < \delta$$ we have $\left|S(f,D_1,\dots, D_{n},P_0,\dots, P_{n})-I\right| < \varepsilon$ for every choice of $$P_1\in D_1$$, $$\dots$$, $$P_{n}\in D_n$$.

The value $$I$$ from the previous definition is called the surface integral of $$f$$ over $$\sigma$$ and is denoted by: $I=\iint_{\sigma}f\,dS.$

Theorem 1 Let $$\sigma$$ be the surface with parametrization $$x=\phi(t,s)$$, $$y=\psi(t,s)$$, $$z=\theta(t,s)$$, where $$t,s\in D$$ for some $$D\subseteq \mathbb R^2$$. Assume that $$\phi$$, $$\psi$$, and $$\theta$$ are differentiable functions, and assume that $$\iint_{\sigma}f\,dS$$ exists. Let us introduce the following notation: \begin{eqnarray*} \overrightarrow R(t,s)&=&\langle \phi(t,s),\psi(t,s),\theta(t,s)\rangle\\ \overrightarrow{R_t}(t,s)&=&\langle \phi_t(t,s), \psi_t(t,s),\theta_t(t,s)\rangle\\ \overrightarrow{R_s}(t,s)&=&\langle \phi_s(t,s),\psi_s(t,s),\theta_s(t,s)\rangle\\ \overrightarrow N(t,s)&=&\overrightarrow{R_t}(t,s)\times \overrightarrow{R_s}(t,s). \end{eqnarray*} The following equality holds: $\iint_{\sigma}f\,dS=\iint_D f\left(\phi(t,s),\psi(t,s),\theta(t,s)\right) \left|\overrightarrow N(t,s)\right|\,dtds.$

Remark: The vectors $$\overrightarrow{R_t}$$ and $$\overrightarrow{R_s}$$ are tangent to the surface, while the vector $$\overrightarrow N$$ is normal to the surface.

Example 1Assume that $$\sigma$$ is the top hemisphere of radius $$3$$. Let $$f:\mathbb R^3\to\mathbb R$$ be the function defined as $$f(x,y,z)=x+y+z$$. Evaluate the integral $$\iint_{\sigma} f\,dS$$.

## Surface integrals of vector fields

Surface integrals of vector fields can be defined only over orientable surfaces. A surface is orientable if it has two sides that can be painted in two different colors. Equivalently, a surface is orientable if we can decide to call one of its sides top’’ and the other bottom.’’ An example of non-orientable surface is a Mobius strip. We will not attempt to provide the most general definition of the orientability. For our purposes it suffices to define the concept for those surfaces that are smooth. Our approach will be using the normal vector: A surface $$\sigma$$ is orientable if there exists a continuous vector field $$\overrightarrow n:\sigma\to\mathbb R^3$$ defined on the surface such that $$\overrightarrow n(P)$$ is perpendicular to $$\sigma$$ at $$P$$ (i.e. to all tangent vectors to $$\sigma$$ at point $$P$$).

Definition: Surface integral of a vector fieldAssume that $$\sigma$$ is an orientable surface and denote by $$\overrightarrow n:\sigma\to\mathbb R^3$$ a continuous normal vector field that fixes its orientation (there are two such vector fields). Assume that one smooth parametrization of $$\sigma$$ is given by $$x=\phi(t,s)$$, $$y=\psi(t,s)$$, $$z=\theta(t,s)$$, where $$(t,s)\in D$$ for some $$D\subseteq \mathbb R^2$$. Let $$\overrightarrow F:\sigma\to\mathbb R^3$$ be a vector field. Let us introduce the following notation: \begin{eqnarray*} \overrightarrow R(t,s)&=&\langle \phi(t,s),\psi(t,s),\theta(t,s)\rangle\\ \overrightarrow{R_t}(t,s)&=&\langle \phi_t(t,s), \psi_t(t,s),\theta_t(t,s)\rangle\\ \overrightarrow{R_s}(t,s)&=&\langle \phi_s(t,s),\psi_s(t,s),\theta_s(t,s)\rangle\\ \overrightarrow N(t,s)&=&\pm\overrightarrow{R_t}(t,s)\times \overrightarrow{R_s}(t,s). \end{eqnarray*} In the last line the choice of $$+$$ or $$-$$ is made so that $$\overrightarrow N(t,s)$$ has the same direction as the chosen orientation vector $$\overrightarrow n$$ at the point $$(\phi(t,s),\psi(t,s),\theta(t,s))$$. We define the surface integral of $$\overrightarrow F$$ over the oriented surface $$\sigma$$ as: $\iint_{\sigma} \overrightarrow F\cdot d\overrightarrow S=\iint_D \overrightarrow F\left(\phi(t,s),\psi(t,s),\theta(t,s)\right)\cdot \overrightarrow N(t,s)\,dtds.$

Remark There are several ways to write the surface integral of a vector field $$\overrightarrow F=\langle P, Q, R\rangle$$ over the surface $$\sigma$$: $\iint_{\sigma} \overrightarrow F\cdot d\overrightarrow S=\iint_{\sigma} \overrightarrow F\cdot \overrightarrow n\,dS=\iint_{\sigma}P\,dydz+Q\,dzdx+R\,dxdy.$

Example 2 Let $$\sigma$$ be the portion of the paraboloid $$z=4-x^2-y^2$$ above the $$xy$$ plane oriented using the downward unit normal vector. Let $$\overrightarrow F=\langle y,x,z\rangle$$. Evaluate the integral $$\iint_{\sigma} \overrightarrow F\cdot d\overrightarrow S$$.

## Practice problems

Problem 1. If $$\overrightarrow F=3\overrightarrow i+y\overrightarrow j$$, and if the surface $$G$$ is given by its parametrization $$\overrightarrow R(s,t)=\langle 9\cos s , 9\sin s, t\rangle$$ for $$0\leq s\leq 2\pi$$ and $$0\leq t\leq 10$$, evaluate $\iint_G \overrightarrow F\cdot \overrightarrow n\,dS,$ where $$\overrightarrow n$$ is the unit normal vector that is pointing away from $$z$$ axis at any point of the surface.

Problem 2. Assume that $$\overrightarrow F$$ is a differentiable vector field and that $$f$$ is a differentiable function in three dimensional space. Assume that $$G$$ is a surface and that $$\overrightarrow n$$ is a continuous unit normal vector field on the surface. Which of the following expressions represents a surface integral of a function?
• (A) $$\displaystyle \iint_G \nabla\overrightarrow F\cdot \overrightarrow n\,dS$$
• (B) $$\displaystyle \iint_G \nabla f\,dS$$
• (C) $$\displaystyle\iint_G\nabla \overrightarrow F\,dS$$
• (D) $$\displaystyle \iint_G\nabla f\cdot \overrightarrow F\cdot \overrightarrow n\,dS$$
• (E) $$\displaystyle \iint_G\nabla f\cdot\overrightarrow F\,dS$$

Problem 3. Find the area of the part of the sphere $$x^2+y^2+z^2=100$$ that lies above the $$xy$$ plane and within the cylinder $$x^2+(y-5)^2=25$$.

Problem 4. Evaluate the surface integral $\iint_G (x^2+y^2+2z)\,dS,$ where $$G$$ is the part of the paraboloid $$z=7-x^2-y^2$$ that lies above the $$xy$$-plane.

Problem 5. Let $$\gamma$$ be the curve in the $$xy$$ plane representing the graph of the function $$f(x)=x^3+2x^2$$ for $$0\leq x\leq 3$$. Let $$G$$ be the surface of revolution obtained by rotation of the curve $$\gamma$$ around the $$y$$ axis. Let $$\overrightarrow F$$ be the vector field given by $$\overrightarrow F=\langle y^2, x+y,z\rangle$$. Evaluate the integral $\iint_G \overrightarrow F\cdot \overrightarrow n\,d S,$ where $$\overrightarrow n$$ is the unit normal vector to the surface $$G$$ that forms an angle smaller than $$90^{\circ}$$ with the $$y$$-axis.