Math2111: Chapter 4: Surface integrals. Section 2: Surface area and surface integrals of scalar fields

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

In the following we assume that the surfaces are smooth, that is, they are assumed to be images of parameterised surfaces $\boldsymbol{\Phi}:D\to\mathbb{R}^3$ for which:

• $D$ is a non-empty, compact and Jordan-measurable subset of $\mathbb{R}^2$;
• the mapping $\boldsymbol{\Phi}$ is one-to-one;
• $\boldsymbol{\Phi}$ is continuously differentiable
• the normal vector $\boldsymbol{n}=\frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v} \neq \boldsymbol{0}$ except possibly at a finite number of points;

(Notice, the condition that ${}D$ is compact can also be replaced by the condition that the surface $S=\{\boldsymbol{\Phi}(u,v): (u,v)\in D\}$ is compact.)

Surface area

In a previous post we discussed parameterised surfaces. Now we calculate the area of parameterised surfaces.

Recall that the area of a parallelogram in $\mathbb{R}^3$ spanned by two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ is given by the Euclidean norm $\|\cdot\|_2$ of the vector obtained by taking the cross product of these two vectors, that is, by $\|\boldsymbol{a}\times \boldsymbol{b}\|_2.$

From the parameterisation $\boldsymbol{\Phi}:D\subset\mathbb{R}^2\to\mathbb{R}^3$ we obtain two tangent vectors $\frac{\partial \boldsymbol{\Phi}}{\partial u}$ and $\frac{\partial \boldsymbol{\Phi}}{\partial v}.$ We can approximate a piece of the surface at some point $\boldsymbol{\Phi}(u_k,v_k)$ ($(u_k,v_k)\in D$) by a parallelogram spanned by the vectors $\frac{\partial \boldsymbol{\Phi}}{\partial u}(u_k,v_k) \Delta u_k$ and $\frac{\partial \boldsymbol{\Phi}}{\partial v}(u_k,v_k) \Delta v_k,$ whose area can be approximated by

$\displaystyle \begin{array}{lr} \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u}(u_k,v_k) \Delta u_k \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u_k,v_k) \Delta v_k \right\|_2 & \\ & \\ = \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u}(u_k,v_k) \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u_k,v_k) \right\|_2 \Delta u_k \Delta v_k. & \end{array}$

By summing over all pieces which approximate the whole surface, i.e. forming the sum

$\displaystyle \sum_{k=1}^N \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u}(u_k,v_k) \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u_k,v_k) \right\|_2 \Delta u_k \Delta v_k,$

and considering the limit when the size of the pieces goes to zero we obtain the integral

$\displaystyle A(S) = \iint_D \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u}(u,v) \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u,v) \right\|_2 \,\mathrm{d} u \,\mathrm{d} v = \int_D \left\|\boldsymbol{n}(u,v) \right\|_2 \,\mathrm{d} u \,\mathrm{d} v.$

(Here, $\boldsymbol{n}$ is the normal vector defined here.) We call $A(S)$ the surface area of the surface ${}S$.

Definition
Let $\boldsymbol{\Phi}:D\subset\mathbb{R}^2\to\mathbb{R}^3$ be a parameterisation of a surface ${}S.$ Then the surface area $A(S)$ of ${}S$ is defined by

$\displaystyle A(S)= \iint_D \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u}(u,v) \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u,v) \right\|_2 \,\mathrm{d} u \,\mathrm{d} v.$

The last formula can also be written as

$\displaystyle \begin{array}{rcl} A(S) &=& \iint_D \|\boldsymbol{n}(u,v)\|_2 \,\mathrm{d} u \,\mathrm{d} v \\ && \\ &=& \iint_D \sqrt{\left(\frac{\partial (Y,Z)}{\partial (u,v)} \right)^2 + \left(\frac{\partial (Z,X)}{\partial (u,v)} \right)^2 +\left(\frac{\partial (X,Y)}{\partial (u,v)} \right)^2} \,\mathrm{d}u\,\mathrm{d} v. \end{array}$

If the surface is a graph of a function $f:D\subset\mathbb{R}^2\to\mathbb{R},$ then $\boldsymbol{n}=\left(-\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1\right)$ and hence the surface of the graph is given by

$\displaystyle A(S)=\iint_D \sqrt{\left(\frac{\partial f}{\partial x} \right)^2 +\left(\frac{\partial f}{\partial y} \right)^2 + 1} \,\mathrm{d} x\,\mathrm{d} y.$

Example
Consider a sphere of radius $R\textgreater 0.$ To calculate its surface area, notice that, because of symmetry, we can calculate the surface area of the upper hemisphere and multiply the result by ${}2$ to obtain the surface area of the whole sphere. The upper hemisphere is given by the equation $x^2+y^2+z^2=R^2,$ where $z\textgreater 0.$ We can set $z = f(x,y)=\sqrt{R^2-x^2-y^2}$ and use the parameterisation of surfaces for functions as shown in Section 1. The parameter domain ${}D$ is in this case $D=\{(x,y)\in\mathbb{R}^2: x^2+y^2\le R^2\}$ and the normal vector is

$\displaystyle \boldsymbol{n}(x,y)= \frac{x}{\sqrt{R^2-x^2-y^2}} \widehat{\boldsymbol{i}} + \frac{y}{\sqrt{R^2-x^2-y^2}} \widehat{\boldsymbol{j}} + \widehat{\boldsymbol{k}}.$

The length of this vector is

$\displaystyle \|\boldsymbol{n}(x,y)\|_2= \frac{R}{\sqrt{R^2-x^2-y^2}}.$

Hence the surface area of the hemisphere (which we shall denote by $A(S/2)$) is given by

$\displaystyle A(S/2)= \iint_D \|\boldsymbol{n}(x,y)\|_2 \,\mathrm{d} x\,\mathrm{d} y = \iint_D \frac{R}{\sqrt{R^2-x^2-y^2}} \,\mathrm{d} x\,\mathrm{d} y.$

The last integral can be evaluated using polar coordinates by which we obtain

$\displaystyle A(S/2)=\int_0^R\int_{0}^{2\pi} \frac{R}{\sqrt{R^2-r^2}} r \,\mathrm{d} \theta \,\mathrm{d} r = 2\pi R \int_0^r \frac{1}{\sqrt{R^2-r^2}} r\,\mathrm{d} r = 2\pi R^2.$

Hence the area of the sphere is given by $\displaystyle A(S) = 2A(S/2) = 4\pi R^2.$ $\Box$

Exercise
Calculate the surface area of a cone parameterised by $x=r\cos\theta,$ $y = r\sin\theta$ and $z=r,$ where $0\le \theta \le 2\pi$ and $0\le r\le 1.$
$\Box$

Scalar surface integrals

We now integrate scalar fields over surfaces. This is in analogy to scalar line integrals considered in Chapter 3, Section 1.

Definition
Let $\boldsymbol{\Phi}:D\subset\mathbb{R}^2\to\mathbb{R}^3$ be a parameterisation of the surface $S=\mbox{Image}(\boldsymbol{\Phi})$ and let $f:S\to\mathbb{R}$ be continuous. Then the integral of ${}f$ over ${}S$ is given by

$\displaystyle \iint_S f \,\mathrm{d} \mathcal{S} = \iint_D f(\boldsymbol{\Phi}(u,v)) \left\|\frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v} \right\|\,\mathrm{d} u\,\mathrm{d} v.$

The last formula can also be written as

$\displaystyle \begin{array}{rcl} \iint_S f \,\mathrm{d} \mathcal{S} &=& \iint_D f(\boldsymbol{\Phi}(u,v)) \|\boldsymbol{n}(u,v)\|_2 \,\mathrm{d} u \,\mathrm{d} v \\ && \\ &=& \iint_D f(\boldsymbol{\Phi}(u,v)) \sqrt{\left(\frac{\partial (Y,Z)}{\partial (u,v)} \right)^2 + \left(\frac{\partial (Z,X)}{\partial (u,v)} \right)^2 +\left(\frac{\partial (X,Y)}{\partial (u,v)} \right)^2} \,\mathrm{d}u\,\mathrm{d} v. \end{array}$

If the surface is the graph of a function $g:D\to\mathbb{R},$ then we also have

$\displaystyle \iint_S f \,\mathrm{d} \mathcal{S} = \iint_D f(x,y,g(x,y)) \sqrt{\left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2 + 1} \,\mathrm{d} x\,\mathrm{d} y .$

Example
Let a surface ${}S$ be given by $z^2=x^2+y^2$ with $0\le z \le 1$ and let $f(x,y,z)=1+z + x^2+y^2.$ Then set $X(r,\theta)=r\cos\theta,$ $Y(r,\theta)=r\sin\theta$ and $Z(r,\theta)=r.$ Then

$\displaystyle \frac{\partial (Y,Z)}{\partial (r,\theta)} = r\cos \theta, \frac{\partial (Z,X)}{\partial (r,\theta)} =-r\sin\theta, \frac{\partial (X,Y)}{\partial (r,\theta)}=r.$

Hence

$\displaystyle \begin{array}{rcl} \iint_S f\,\mathrm{d} \mathcal{S} &= & \int_{0}^1 \int_{0}^{2\pi} f(X(r,\theta), Y(r,\theta),Z(r,\theta)) \sqrt{r^2\sin^2\theta+r^2\cos^2\theta+r^2}\,\mathrm{d}\theta\,\mathrm{d} r \\ && \\ &=& \int_0^1 \int_0^{2\pi} (1+r+r^2) r \sqrt{2}\,\mathrm{d}\theta\,\mathrm{d} r = 2\sqrt{2}\pi \frac{13}{12}. \end{array}$

$\Box$

Exercise
The surface in the previous example is the graph of a function. Use this to parameterise the surface and calculate the scalar surface integral using this approach.
$\Box$

Surface integrals of scalar valued functions over graphs

Suppose the surface ${}S$ is the graph of a function $z=g(x,y)$ defined on a domain $D\subset\mathbb{R}^2.$ Then we can use the parameterisation $\boldsymbol{\Phi}(x,y)=x\widehat{\boldsymbol{i}} + y\widehat{\boldsymbol{j}} + g(x,y) \widehat{\boldsymbol{k}}.$ Then the normal vector is $\boldsymbol{n}(x,y)= -\frac{\partial g}{\partial x} \widehat{\boldsymbol{i}} -\frac{\partial g}{\partial y} \widehat{\boldsymbol{j}} + \widehat{\boldsymbol{k}}.$ Hence $|\boldsymbol{n} \cdot \widehat{\boldsymbol{k}}| = 1$ and therefore

$\displaystyle \iint_S f \,\mathrm{d}\mathcal{S}= \iint_D f \|\boldsymbol{n}\|\,\mathrm{d} A = \iint_D f \frac{\|\boldsymbol{n}\|}{|\boldsymbol{n}\cdot \widehat{\boldsymbol{k}}|}\,\mathrm{d} A = \iint_D \frac{f}{|\widehat{\boldsymbol{n}}\cdot \widehat{\boldsymbol{k}}|} \,\mathrm{d} A,$

where $\widehat{\boldsymbol{n}}$ is the unit normal vector to the surface ${}S.$

Example
Let a surface ${}S$ be given by $2x+y-2z=1$ and $x,y\ge 0$ and $z \le 0.$ Let $f(x,y,z) = y.$ Then a normal vector to the surface is $\boldsymbol{n} = (2,1,-2).$ Since $\|\boldsymbol{n}\|=3,$ the unit normal vector is $\widehat{\boldsymbol{n}}=(2/3,1/3,-2/3).$ Hence $|\widehat{\boldsymbol{n}} \cdot \widehat{\boldsymbol{k}}| = 2/3$. We can describe the surface as a graph of the function $z = g(x,y)=x+y/2-1/2,$ where the domain is $D=\{(x,y): 0 \le x \le 1/2, 0 \le y \le 1-2x \}.$ Therefore

$\displaystyle \iint_S f \,\mathrm{d} \mathcal{S}=\iint_D \frac{f}{|\widehat{\boldsymbol{n}}\cdot \widehat{\boldsymbol{k}}|} \,\mathrm{d} A = \int_0^{1/2} \int_0^{1-2x} 3y/2 \,\mathrm{d} y \,\mathrm{d} x = \frac{1}{8}.$

$\Box$

Applications

Scalar surface integrals can be used to calculate mass, center of mass and moments of inertia of thin shells. Let $\delta$ be the density function of a very thin shell.

• Mass $M =\iint_S \delta \,\mathrm{d} \mathcal{S}$
• Center of mass $\overline{x} = \frac{1}{M}\iint_S x \delta \,\mathrm{d} \mathcal{S},$ $\overline{y}=\frac{1}{M} \iint_S y \delta \,\mathrm{d} \mathcal{S}$ and $\overline{z} = \frac{1}{M} \iint_S z \delta \,\mathrm{d} \mathcal{S}.$
• Moment of inertia $I_x=\iint_S (y^2+z^2)\delta \,\mathrm{d} \mathcal{S},$ $I_y=\iint_S (x^2+z^2)\delta \,\mathrm{d} \mathcal{S}$ and $I_z=\iint_S (x^2+y^2)\delta \,\mathrm{d} \mathcal{S}$