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\(\newcommand{\R}{\mathbb R}\)
\(\Leftarrow\) \(\Uparrow\) \(\Rightarrow\)
Suppose that \(S\) is a surface in \(\R^3\) parametrized by a function \(\mathbf G:T\to S\), where \(T\) is a measurable subset of \(\R^2\), \(\mathbf G\) is \(C^1\) and injective, and that \(\left\{\partial_1 \mathbf G, \partial_2 \mathbf G\right\}\) are linearly independent at every point in \(T\), except possibly on a set of zero content.
If \(|\partial_1 \mathbf G \times \partial_2 \mathbf G|\) is integrable on \(T\), then we define the surface area of \(S\) as \[ \iint_T |\partial_1 \mathbf G \times \partial_2 \mathbf G|\, dA \]We also sometimes say or write “integral” instead of “surface integral” when referring to the above integrals, if it is clear that we are integrating over a surface.
Note that the area of \(S\) equals \(\iint_S 1\, dA\).
Although we have used a parametrization in the above definitions, we will see below that, exactly as with line integrals,
Let \(S\) be the set \(\left\{ (x,y,z)\in \R^3 : x^2+y^2+z^2 = 1, \ z> 0\right\}\). Compute the area of \(S\), as well as the integrals \[ \iint_S z\, dA, \quad\text{ and }\quad \iint_S (0,0,z)\cdot \mathbf n \, dA, \] with \(\mathbf n\) oriented upwards (i.e. \(\mathbf n \cdot [0,0,1]>0\)).
Solution.
First we parametrize \(S\). We can do this by spherical coordinates with \(r=1\), \[
\mathbf x = \mathbf G(\theta,\phi)= (\cos\theta \sin\phi , \sin\theta\sin\phi ,\cos \phi), \qquad 0\le\theta\le 2\pi, \ \ 0\le \phi\le \pi/2 .
\] Now we compute the integrals. \[\begin{align*}
\partial_\theta \mathbf G &= ( -\sin\theta \sin\phi , \cos\theta\sin\phi ,0) \\
\partial_\phi\mathbf G &=( \cos\theta \cos\phi , \sin\theta\cos\phi ,-\sin \phi),\quad\text{ and thus } \\
\partial_\theta \mathbf G\times \partial_\phi \mathbf G &=(-\cos\theta\sin^2\phi, -\sin\theta\sin^2\phi, -\cos\phi\sin \phi)
\end{align*}\] Then a short computation shows that \[
|\partial_\theta \mathbf G\times \partial_\phi \mathbf G| = |\sin \phi| = \sin \phi
\] since \(\sin \phi\ge 0\) for \(\phi\in [0,\pi/2]\). It follows that \[\begin{align*}
&\text{area}(S) = \int_0^{2\pi}\int_0^{\pi/2} \sin \phi \, d\phi\, d\theta = \cdots = 2\pi \\
&\iint_S z \, dA =\int_0^{2\pi}\int_0^{\pi/2} \cos\phi \sin\phi\, d\phi \, d\theta = \pi.
\end{align*}\] The substitution \(u = \sin \phi\) is helpful in evaluating the second integral.
When \(\left\{\partial_1 \mathbf G, \partial_2 \mathbf G\right\}\) are linearly independent (as we typically require, except possibly for a set of zero content), then
As a result of these considerations, \[ \frac{ \partial_1 \mathbf G \times \partial_2 \mathbf G } {|\partial_1 \mathbf G \times \partial_2 \mathbf G|} \qquad\text{ and }\qquad \frac{ -\partial_1 \mathbf G \times \partial_2 \mathbf G } {|\partial_1 \mathbf G \times \partial_2 \mathbf G|} = \frac{ \partial_2 \mathbf G \times \partial_1 \mathbf G } {|\partial_2 \mathbf G \times \partial_1 \mathbf G|} \] are unit normal vectors to \(S\) — unit vectors that are normal to \(S\) at the point \(\mathbf G(u,v)\).
We therefore sometimes write \(\dfrac{\partial_1 \mathbf G \times \partial_2 \mathbf G} {|\partial_1 \mathbf G \times \partial_2 \mathbf G|}\) as \(\mathbf n(\mathbf G(\mathbf u))\), where “\(\mathbf n\)” stands for “normal”.
With this notation, for any vector field \(\mathbf F\), the surface integral of the real-valued function \(\mathbf F\cdot \mathbf n\) over \(S\) is \[\begin{align} \iint_S \mathbf F\cdot \mathbf n \,dA &= \iint_T \mathbf F(\mathbf G(\mathbf u))\cdot \left(\partial_1 \mathbf G(\mathbf u) \times \partial_2 \mathbf G(\mathbf u)\right)\, du\, dv\nonumber \\ &= \iint_T \mathbf F(\mathbf G(\mathbf u))\cdot \mathbf n(\mathbf G(\mathbf u)) \left| \partial_1 \mathbf G \times \partial_2 \mathbf G \right| \, du\, dv\nonumber \\ \end{align}\] For this reason, the notation \(\iint_S \mathbf F\cdot \mathbf n\, dA\) for a surface integral of a vector field is entirely consistent with the notation for the surface integral of a real-valued function.
So far we have discussed the direction of \(\partial_1 \mathbf G \times \partial_2 \mathbf G\). What about its magnitude? Properties of the cross product imply that its magnitude is exactly the area of the parallelogram in \(\R^3\) with sides \(\partial_1 \mathbf G\) and \(\partial_2 \mathbf G\). This leads to a non-rigorous but suggestive interpretation, which is that \[\begin{align} | \partial_1 \mathbf G \times \partial_2 \mathbf G|\,du\,dv & = | (\partial_1 \mathbf G du) \times (\partial_2 \mathbf G dv)|\nonumber \\ \end{align}\] regarding this as the area of the infinitesimal parallelogram with sides \(\partial_1 \mathbf G du\) and \(\partial_2 \mathbf G dv\).
Thus, it is traditional to imagine that when we integrate \(\iint_R | \partial_1 \mathbf G \times \partial_2 \mathbf G|\,du\,dv\), we are “adding up the areas of the infinitely many infinitesimal parallelograms that make up the surface \(S\)”. This can be viewed as a sort of justification for our defintion of area. One can also make similar arguments with rectangles of very small but but finite size. Other justifications will be presented below.
Let \(S\) be the boundary of the set \(\left\{ (x,y,z)\in \R^3 : x^2+y^2\le 1, |z|\le 1\right\}\). Compute \(\iint_S \mathbf F\cdot \mathbf n\, dA\) for \(\mathbf F = (x^2z,y,0)\), and with \(\mathbf n\) oriented outwards.
Solution.
A feature of this problem is that \(\partial S\) is not a smooth surface everywhere. However, it is a piecewise smooth surface, that is, a union of smooth surfaces: \[
\begin{aligned}
S_{top} &= \left\{(x,y,z)\in \R^3 : \ x^2+y^2\le 1, \ z = 1\right\} \\
S_{side} &= \left\{(x,y,z)\in \R^3 : \ x^2+y^2 = 1, \ |z|\le 1\right\} \\
S_{bottom} &= \left\{(x,y,z)\in \R^3 : \ x^2+y^2\le 1, \ z = - 1\right\}
\end{aligned}
\] that intersect at sets of lower dimension. For example, \(S_{top}\cap S_{side} = \left\{(x,y,z) : x^2+y^2 = 1, z = 1\right\}\) which is a smooth curve in \(\R^3\), hence has zero area. In cases like this, we simply integrate along each of the surfaces and add up the results.
Integral over \(S_{top}\).. With some practice you can see right away that this integral must equal \(0\), because we can see geometrically that \(\mathbf n = (0,0,1)\) on \(S_{top}\), and hence that \(\mathbf F\cdot \mathbf n = 0\) there, for our specific \(\mathbf F\). But for practice, let’s do it:
Integral over \(S_{bottom}\). For exactly the same reason, \[ \iint_{S_{bottom}} \mathbf F\cdot \mathbf n \, dA = 0. \] For practice, let’s not do it. It’s just as important to practice avoiding computations as doing computations.
Integral over \(S_{side}\)..
The above formulas simplify if \(S\) is the graph of some function \(\phi\), i.e. \(S = \left\{ (x,y, \phi(x,y) ): (x,y)\in T\right\}\). Then we can use the parametrization \[ \mathbf x = (u , v , \phi(u,v)) = \mathbf G(u,v), \qquad (u,v)\in T. \] With this choice, \[ \partial_u\mathbf G = (1, 0, \partial_u \phi), \qquad \partial_v\mathbf G = ( 0,1, \partial_v \phi), \qquad \partial_u\mathbf G\times \partial_v\mathbf G=(-\partial_u \phi, -\partial_v\phi, 1). \] and thus \(|\partial_u\mathbf G\times \partial_v\mathbf G| = \sqrt {1+|\nabla\phi|^2}\). So we get the simpler formulas \[ \text{area}(S) = \iint_T \sqrt {1+|\nabla\phi|^2}\, dA \] \[ \iint_S f\, dA = \iint_T f(u,v,\phi(u,v)) \sqrt {1+|\nabla\phi|^2} \, dA \] and \[ \iint_S \mathbf F\cdot \mathbf n \, dA = \iint_T \mathbf F(u,v,\phi(u,v)) \cdot (-\partial_u\phi, -\partial_v\phi, 1) \, dA \] if \(S\) is oriented with the unit normal pointing upwards; for the downward unit normal, one would have to multiply by \(-1\).
Since \(u=x\) and \(v=y\), these formulas are often written in terms of variables \((x,y)\) rather than \((u,v)\).
Similarly to the line integral, we will show that the integral of a real-valued function \(f\) over a surface \(S\) is independent of the parametrization.
Suppose that \(S\) is a surface parametrized by \[ (x,y,z) = \mathbf G(u,v) \ \ \ \text{ for }(u,v) \in T, \] and that \(\varphi:W\to T\) is a function that is one-to-one, onto, of class \(C^1\), and with \(C^1\) inverse. Here both \(W\) and \(T\) are supposed to be measurable subsets of \(\R^2\). Let us say that the variables in \(W\) are called \((s,t)\), so that the \((s,t)\) and \((u,v)\) variables are related by \((u,v) = \varphi(s,t)\).
Now define \(\mathbf H(s,t) = \mathbf G\circ\varphi(s,t)\). Then \(S\) is also parametrized by \(\mathbf x = \mathbf H(s,t), (s,t)\in W\).
Our assumptions about \(\varphi\) imply that \(\det D\varphi(s,t)\ne 0\) for all \((s,t)\in W\). Also, by a short computation that uses the chain rule, one can check that
\[\begin{equation}\label{cvsurf} \frac{\partial \mathbf H}{\partial s}\times \frac{\partial \mathbf H}{\partial t} = \left(\frac{\partial \mathbf G}{\partial u}\times \frac{\partial \mathbf G}{\partial v}\right) \cdot \det D\varphi \end{equation}\]
where derivatives of \(\mathbf H, \varphi\) are evalated at \(s,t\) and derivatives of \(\mathbf G\) are evaluated at \((u,v) = \varphi(s,t)\). It follows that
\[ \begin{aligned} \det D\varphi >0 & \text{ everywhere in }W \\ &\ \iff \ \frac{\partial \mathbf H}{\partial s}\times \frac{\partial \mathbf H}{\partial t} \text{ and } \frac{\partial \mathbf G}{\partial u}\times \frac{\partial \mathbf G}{\partial v}\text{ point in the same direction} \\ &\ \iff \ \text{ parametrizations }\mathbf G\text{ and }\mathbf H\text{ induce the same orientation}, \end{aligned} \] where the orientation induced by a parametrization \(\mathbf G\) is the orientation of \(\mathbf n = \partial_u\mathbf G\times \partial_v \mathbf G/|\partial_u\mathbf G\times \partial_v \mathbf G|\).
The special case \(f=1\) reduces to the area of a surface \(S\) is independent of the parametrization
This is nice to know, because if our definition of “area” depended on the parametrization, then it would probably not be a good definition.
For integrals of vector fields, as already discussed, we have to pay attention to the orientation, which is the choice of the direction in which the unit normal \(\mathbf n\) points, as we have already seen in examples. That is, the integral of a vector field \(\mathbf F\) over a surface \(S\) depends on the orientation of \(S\) but is otherwise independent of the parametrization. In fact, changing the orientation of a surface (which amounts to multiplying the unit normal \(\mathbf n\) by \(-1\), changes the sign of the surface integral of a vector field.
Because of the above facts, if we want to integrate a real-valued function over a surface, or (in particular) to find the area of a surface, we do not need to know how the surface is oriented. But to compute the surface integral of a vector field, we need to specify the orientation of the surface. So if someone asks you to compute \(\iint_S \mathbf F\cdot \mathbf n\, dA\), it is their job to specify not only the vector field \(\mathbf F\) and the surface \(S\), but also the orientation of the unit normal \(\mathbf n\). One that is done, we can speak unambiguously about \(\iint_S \mathbf F\cdot \mathbf n \, dA\), without having to say precisely which parametrization we have in mind.
Surface area is a subtler concept than volume. If you are asked to measure the volume of a physical object, it is (in principle) easy to do: dunk the object in a bucket of water, and measure how much water it displaces.
Measuring surface area is much more difficult. How might you do it?
One possibility is to paint the object on one side with a thin coat of paint. Then the area of the surface should be proportional to the volume of paint used, which we can measure. For example, if we are able to apply paint with a uniform thickness of exactly \(h\) cm, (a realistic number might be \(h = 0.1\, cm\)) and if we use \(V\) \(cm^3\) of paint, then the area should be about \[ \frac{ V \, cm^3}{h \, cm} = \frac V h \, cm^2. \]
If we coat \(S\) with a layer of paint of thickness \(h\), the painted sphere occupies the region \[ S_h = \left\{ \mathbf x\in \R^3 : 1 \le |\mathbf x| \le 1+h\right\}. \] The volume is easily computed and equals \[ \text{Vol}(S_h) = \frac 43 \pi [(1+h)^3 - 1^3] = \frac 43\pi(3h +3h^2+h^3). \] So the area of the sphere should be approximately \[ \text{area}(S) \approx \frac {\frac 43\pi(3h^2 +3h+h^3)}h = 4\pi + 4\pi h + \frac 43\pi h^2. \] If \(h\) is small, this is very close to the number you have probably computed in first-year calculus or elsewhere, \(4\pi\).
If we had (somehow) painted \(S\) on the inside, we would have gotten a different approximation, area\((S)\approx 4\pi - 4\pi h + \frac 43 \pi h^2\). But these two different approximations would yield the same number for the area if we take the limit \(h\to 0\).These considerations suggest a way of computing area of a surface \(S\) that is a mathematical idealization of the procedure of measuring the volume of a thin coat of paint.
Choose an orientation \(\mathbf n\) for the surface. (It will not matter which you choose.)
For \(h>0\), define \[\begin{equation}\label{Sh.def} S_h = \left\{ \mathbf x + s \mathbf n : \mathbf x \in S, 0 \le s \le h \right\}. \end{equation}\] This is the set of all points on the side of the surface in the \(\mathbf n\) direction, that are a distance as most \(h\) from the surface, or if you prefer, the very thin region occupied by a coat of paint of thickness \(h\).
Then we expect that \[ \text{area}(S) = \lim_{h\to 0}\frac{\text{Vol}(S_h)}{h}. \] As the example of the sphere suggests, this procedure should yield a result that does not depend on the orientation of \(\mathbf n\).
In fact this is true:
We will never use the theorem; its only role for us is as a motivation/justification for the definition of area. You may find it to be a more convincing justification than the hand-waving arguments about adding up areas on infinitesimal parallelograms alluded to earlier.
The proof is a little too involved for this class, but we can at least describe the outline.
For \(s>0\) define \[ \widetilde{\mathbf G} (u,v,s) = \mathbf G (u,v) + s \mathbf n (\mathbf G (u,v)), \] where \[ \mathbf n(\mathbf G (u,v)) = \frac{\partial_u\mathbf G \times \partial_v \mathbf G}{\left|\partial_u\mathbf G \times \partial_v \mathbf G \right|}(u,v). \] Also, let \(R_h = \left\{(u,v,s) : (u,v)\in R, 0\le s \le h\right\}\). Then
Check that \(\widetilde{\mathbf G}\) is a transformation of \(R_h\) onto \(S_h\) (that is, one-to-one and onto, with \(C^1\) inverse.)
Then in principle we can compute the volume of \(S_h\) using the Change of Variables Theorem, leading to \[ \text{Vol}(S_h) = \iint_{R_h} |\det \widetilde{\mathbf G} (u,v,s)|\, du\,dv\,ds. \]
Next, one can prove that \[ \lim_{h\to 0}|\det \widetilde{\mathbf G} (u,v,s)| = \left|\partial_u\mathbf G \times \partial_v\mathbf G \right|(u,v). \]
Once this is known, then completing the proof of the theorem is relatively straightforward.
In all these questions, the most important part is to set up the integrals correctly, since this is the new part. We already know (or don’t know, as the case may be) how to evaluate them.
Compute the area of the following surfaces.
\(S = \left\{ (x,y,z)\in \R^3 : x^2+y^2 = z^2, \ 1\le z \le 2\right\}\).
\(S = \left\{(x,y,z) \in \R^3 : x^2+y^2+z^2 = 4, (x-1)^2+ y^2 \le 1\right\}\).
\(S\) is the boundary of the set of points in the first octant of \(\R^3\) where \(x+y+z\le 1\). In principle you can do this by elementry geometry, but using calculus might make things easier.
\(S = \left\{(x,y,z) \in \R^3 : \sqrt{y^2+z^2} = 1-x^2\right\}\).
Derive a formula for the area of a surface \(S\) of the form \(S = \left\{(x,y,z)\in \R^3 : \sqrt{x^2+y^2} = f(z), a\le z\le b\right\}\).
Evaluate the following integrals.
Find the centroid of the set \(\left\{(x,y,z)\in \R^3 : x^2+y^2 + 4z^2 = 4, z\ge 0\right\}\). (The centroid of a set is defined in Section 4.4. When computing the centroid of a surface, you should replace \(dV\) by \(dA\).)
Compute \(\iint_S xy \, dA\) when \(S\) is the portion of the unit sphere in \(\R^3\) that sits in the first octant (thatis, all coordinates are nonnegative.)
Compute the following integrals.
Let \(S\) be a surface of the form \(S = \left\{ (x,y,z) : (x,y)\in W, z= \phi(x,y)\right\}\), with the unit normal oriented upwards, where \(W\) is a measurable subset of \(\R^2\) and \(\phi\) is \(C^1\). Write down a general formula for \(\iint_S \mathbf F\cdot \mathbf n \, dA\), where \(\mathbf F(x,y,z) = (0,0,z)\).
Let \(S\) be the unit sphere in \(\R^3\), with \(\mathbf n\) oriented outwards, and compute \(\iint_S \mathbf F\cdot \mathbf n\, dA\) where \(\mathbf F = (x,y,z)\).
Let \(S\) be the ellipse \((\frac x a)^2 + (\frac yb)^2 + (\frac zc)^2 = 1\), with \(\mathbf n\) oriented inwards, and compute \(\iint_S \mathbf F\cdot \mathbf n\, dA\) for \(\mathbf F = (-a^2y,b^2x,z^2)\).
Let \(S = \left\{(x,y,z)\in \R^3 : (x,y)\in [0,1]^2, z = xy \right\}\), with \(\mathbf n\) oriented downwards, and compute \(\iint_S \mathbf F\cdot \mathbf n\, dA\) for \(\mathbf F = (zx, z^2, xy )\).
Prove \(\eqref{cvsurf}\).
Use \(\eqref{cvsurf}\) to prove Theorem 1.
\(\Leftarrow\) \(\Uparrow\) \(\Rightarrow\)
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