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Section 7.2 Unoriented surface integrals

We define unoriented surface integrals of functions along parametric surfaces in \(\mathbb{R}^3\text{.}\) As a special case, we study how to calculate the surface area of a surface in \(\mathbb{R}^3\text{.}\)

Subsection 7.2.1 Unoriented surface integrals

In the previous section we defined unoriented line integrals over parametric curves in \(\mathbb{R}^n\text{.}\) Those cannot be defined as integrals of differential forms, as the integrals are independent of a choice of orientation on the curve. We can similarly define unoriented surface integrals, to which we now turn to. Those will be useful to calculate quantities associated to parametric surfaces in \(\mathbb{R}^3\) that should not depend on a choice of orientation, such as the surface area of the surface.

We proceed in a way similar to what we did for line integrals, by first recalling our construction of oriented surface integrals using differential forms. Let \(\alpha: D \to \mathbb{R}^3\) be a parametric surface, with \(\alpha(u,v) = (x(u,v), y(u,v), z(u,v) )\text{.}\) We defined the oriented surface integral of a two-form \(\omega\) along \(\alpha\) via pullback to \(D\) (Definition 5.6.1). In terms of the vector field \(\mathbf{F} = (f_1,f_2, f_3)\) associated to the two-form \(\omega\text{,}\) the surface integral can be written as

\begin{equation*} \iint_D (\mathbf{F}(\alpha(u,v)) \cdot \mathbf{n})\ dA, \end{equation*}

where the normal vector \(\mathbf{n}\) is

\begin{equation*} \mathbf{n} = \mathbf{T}_u \times \mathbf{T}_v, \qquad \mathbf{T}_u = \frac{\partial \alpha}{\partial u}, \qquad \mathbf{T}_v = \frac{\partial \alpha}{\partial v}. \end{equation*}

The orientation is now encapsulated in the choice of normal vector. But as was the case for oriented line integrals, the normal vector here contains more information than just the orientation, as it is not normalized. We define the unit normal vector as

\begin{equation*} \hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{\mathbf{T}_u \times \mathbf{T}_v}{|\mathbf{T}_u \times \mathbf{T}_v|}. \end{equation*}

We can then rewrite the surface integral as

\begin{equation*} \iint_D (\mathbf{F}(\alpha(u,v)) \cdot \hat{\mathbf{n}} ) |\mathbf{T}_u \times \mathbf{T}_v|\ dA = \iint_D (\mathbf{F}(\alpha(u,v)) \cdot \hat{\mathbf{n}} )\ dS, \end{equation*}

where we defined the “surface element”

\begin{equation*} dS = |\mathbf{T}_u \times \mathbf{T}_v|\ dA. \end{equation*}

With this rewriting, we can think of the surface integral as integrating the expression \((\mathbf{F}(\alpha(u,v)) \cdot \hat{\mathbf{n}} )\text{,}\) which is a function of the parameters \((u,v)\) that depends on the choice of orientation via the unit normal vector \(\hat{\mathbf{n}}\text{.}\)

Unoriented surface integrals are then naturally defined by replacing the orientation-dependent function \((\mathbf{F}(\alpha(u,v)) \cdot \hat{\mathbf{n}} )\) by an arbitrary function \(f(\alpha(u,v))\) that does not depend on a choice of orientation on the surface. We obtain the following definition of unoriented surface integrals.

Definition 7.2.1. Unoriented surface integrals.

Let \(\alpha: D \to \mathbb{R}^3\) be a parametric surface, with \(\alpha(u,v) = (x(u,v), y(u,v), z(u,v) )\text{,}\) and image surface \(S = \alpha(D) \subset \mathbb{R}^3\text{.}\) The tangent vectors are

\begin{equation*} \mathbf{T}_u = \frac{\partial \alpha}{\partial u}, \qquad \mathbf{T}_v = \frac{\partial \alpha}{\partial v}. \end{equation*}

Let \(f: S \to \mathbb{R}\) be a continuous function. We define the unoriented integral of \(f\) along the surface \(S\) to be

\begin{equation*} \iint_S f\ dS = \iint_D f(\alpha(u,v)) |\mathbf{T}_u \times \mathbf{T}_v|\ dA. \end{equation*}

A calculation similar to the proof of Lemma 5.6.3 shows that unoriented surface integrals are invariant under arbitrary reparametrizations of the surface \(S \subset \mathbb{R}^3\text{,}\) regardless of whether the orientation is preserved or not. This is why we wrote

\begin{equation*} \iint_S f\ dS \end{equation*}

to denote the unoriented surface integral along the surface \(S\text{,}\) as the integral does not depend on how we parametrize the surface.

Evaluate the surface integral

\begin{equation*} \iint_S x^2 z\ dS \end{equation*}

over the cone \(z^2=x^2+y^2\) above the \((x,y)\)-plane and below the plane \(z=1\text{.}\)

We first parametrize the surface as \(\alpha:D \to \mathbb{R}^3\) with

\begin{equation*} D = \{ (r,\theta) \in \mathbb{R}^2\ |\ r \in [0,1], \theta \in [0,2 \pi] \} \end{equation*}

and

\begin{equation*} \alpha(r,\theta) = (r \cos(\theta), r \sin(\theta), r). \end{equation*}

To evaluate the surface integral, we need to calculate the surface element \(dS\text{.}\) The tangent vectors are

\begin{equation*} \mathbf{T}_r = (\cos(\theta), \sin(\theta), 1), \qquad \mathbf{T}_\theta = (- r\sin(\theta), r \cos(\theta), 0). \end{equation*}

The cross product is (we note here that the order does not matter, as the integral is unoriented; we will calculate the norm of the cross product afterwards, which is the same regardless of whether we take \(\mathbf{T}_r \times \mathbf{T}_\theta\) or \(\mathbf{T}_\theta \times \mathbf{T}_r\)):

\begin{equation*} \mathbf{T}_r \times \mathbf{T}_\theta = ( - r \cos(\theta), -r \sin(\theta), r). \end{equation*}

Its norm is

\begin{equation*} |\mathbf{T}_r \times \mathbf{T}_\theta| = \sqrt{r^2 \cos^2(\theta) + r^2 \sin^2(\theta) + r^2} = \sqrt{2} r, \end{equation*}

where we used the fact that \(\sqrt{r^2} = r\) since \(r \in [0,1]\) and hence is positive.

The unoriented surface integral then becomes

\begin{align*} \iint_S x^2 z \ dS =\amp \iint_D (r \cos(\theta) )^2 (r) (\sqrt{2} r)\ dA\\ =\amp \sqrt{2} \int_0^1 \int_0^{2 \pi} r^4 \cos^2(\theta) d\theta dr\\ =\amp \sqrt{2} \pi \int_0^1 r^4\ dr\\ =\amp \frac{\sqrt{2} \pi}{5}. \end{align*}

Subsection 7.2.2 Surface area of a parametric surface in \(\mathbb{R}^3\)

Just as evaluating the unoriented line integral of the constant function \(f=1\) gave the arc length of the parametric curve, evaluating the unoriented surface integral of the constant function \(f=1\) gives the surface area of the parametric surface.

Definition 7.2.3. Surface area of a parametric surface in \(\mathbb{R}^3\).

Let \(\alpha: D \to \mathbb{R}^3\) be a parametric surface, with \(\alpha(u,v) = (x(u,v), y(u,v), z(u,v) )\text{,}\) and image surface \(S = \alpha(D) \subset \mathbb{R}^3\text{.}\) The surface area of \(S\) is given by the unoriented surface integral

\begin{equation*} \iint_S\ dS = \iint_D |\mathbf{T}_u \times \mathbf{T}_v|\ dA. \end{equation*}

The justification for this definition of the surface area comes from the standard slicing process of integral calculus, as usual. Consider a small rectangle at position \((u,v)\) within the domain \(D\text{,}\) and with sides of length \(du\) and \(dv\text{.}\) The area of this small rectangle is \(dA = du dv\text{.}\) The parametrization \(\alpha\) maps the small rectangle to a small region \(dS\) within the surface \(S\text{.}\) This region is now curved, and not necessarily rectangular. However, one can approximate its area by replacing it by the parallelogram in the tangent plane at the point \(\alpha(u,v)\) spanned by the vectors \(d u \mathbf{T}_u\) and \(dv \mathbf{T}_v\text{.}\) The area of this parallelogram is \(dS = |\mathbf{T}_u \times \mathbf{T}_v| du dv\text{.}\) Finally, we sum over all small regions within the domain \(D\text{,}\) and take a limit of an infinite number of infinitesimal regions, which turns the approximate calculation of the surface area into an exact one. The result of this limit process is the double integral

\begin{equation*} \iint_D |\mathbf{T}_u \times \mathbf{T}_v|\ dA. \end{equation*}

Admittedly, the justification is a little bit hand-wavy here, but it can be made precise.

This also justifies our general construction of unoriented surface integrals in Definition 7.2.1: all that we add to the slicing process is a function on the surface \(S\) that we evaluate at each point \(\alpha(u,v)\) on \(S\) before summing over slices to turn the calculation into a double integral.

Find the surface area of the part of the plane \(x+2y+z = 1\) that lies within the cylinder \(x^2+y^2 = 4\text{.}\)

As we know that the region will be within the cylinder \(x^2+y^2=4\text{,}\) we use polar coordinates to parametrize the surface. We write down the parametrization \(\alpha: D \to \mathbb{R}^3\) with

\begin{equation*} D = \{(r, \theta) \in \mathbb{R}^2\ |\ r \in [0,2], \theta \in [0,2\pi] \} \end{equation*}

and

\begin{equation*} \alpha(r,\theta) = (r \cos(\theta), r \sin(\theta), 1 - r \cos(\theta) - 2 r \sin(\theta) ). \end{equation*}

We need to find the surface element \(dS\text{.}\) The tangent vectors are

\begin{equation*} \mathbf{T}_r = (\cos(\theta), \sin(\theta), - \cos(\theta) - 2 \sin(\theta) ), \qquad \mathbf{T}_\theta = (- r \sin(\theta), r \cos(\theta), r \sin(\theta) - 2 r \cos(\theta) ). \end{equation*}

The cross product is

\begin{equation*} \mathbf{T}_r \times \mathbf{T}_\theta = (r, 2r, r), \end{equation*}

whose norm is

\begin{equation*} |\mathbf{T}_r \times \mathbf{T}_\theta| = \sqrt{r^2 + 4 r^2 + r^2} = \sqrt{6} r, \end{equation*}

where we used the fact that \(\sqrt{r^2} = r\) since \(r \in [0,2]\) and hence is positive.

Finally, we evaluate the surface integral of \(dS\) to get the surface area of \(S\text{:}\)

\begin{align*} \iint_S \ dS =\amp \iint_D |\mathbf{T}_r \times \mathbf{T}_\theta|\ dA\\ =\amp \sqrt{6} \int_0^2 \int_0^{2\pi} r \ d\theta dr\\ =\amp 2 \sqrt{6} \pi \int_0^2 r\ dr\\ =\amp 4 \sqrt{6} \pi. \end{align*}

Exercises 7.2.3 Exercises

1.

Find the surface area of the part of the surface \(z=4 - 2 x^2 + y\) over the triangle with vertices \((0,0), (2,0), (2,2)\) in the \((x,y)\)-plane.

Solution.

First, we notice that the triangle in the \((x,y)\)-plane can be realized as the \(x\)-supported region \(x \in [0,2]\text{,}\) \(0 \leq y \leq x\text{.}\) We can then parametrize the surface as \(\alpha:D \to \mathbb{R}^3\) with

\begin{equation*} D = \{(u,v) \in \mathbb{R}^2\ |\ u \in [0,2], 0 \leq v \leq u \} \end{equation*}

and

\begin{equation*} \alpha(u,v) = (u,v,4 - 2 u^2 + v). \end{equation*}

The tangent vectors are

\begin{equation*} \mathbf{T}_u = (1,0,- 4 u), \qquad \mathbf{T}_v = (0,1,1). \end{equation*}

The cross product is

\begin{equation*} \mathbf{T}_u \times \mathbf{T}_v = ( 4u, -1, 1). \end{equation*}

Its norm is

\begin{equation*} |\mathbf{T}_u \times \mathbf{T}_v| = \sqrt{16 u^2 + 1 + 1} = \sqrt{2} \sqrt{8 u^2 + 1}. \end{equation*}

The surface area becomes

\begin{align*} \iint_S\ dS =\amp \sqrt{2} \int_{0}^2 \int_{0}^u \sqrt{8 u^2+1}\ dv du\\ =\amp \sqrt{2} \int_0^2 u \sqrt{8 u^2+1}\ du\\ =\amp \frac{\sqrt{2}}{16} \int_1^{33} \sqrt{t}\ dt\\ =\amp \frac{\sqrt{2}}{24} (33^{3/2} - 1)\\ =\amp \frac{11 \sqrt{66}}{8} - \frac{\sqrt{2}}{24}. \end{align*}

where we did the substitution \(t = 8u^2+1\text{.}\)

2.

Show that the surface area of the lateral surface of a circular cone with radius \(R\) and height \(H\) is \(A = \pi R \sqrt{R^2 + H^2}\text{.}\)

Solution.

The equation of a circular cone with radius \(R\) and height \(H\) (with apex at the origin) is

\begin{equation*} \frac{R^2}{H^2} z^2 = x^2+y^2, \qquad z \geq 0. \end{equation*}

We parametrize the cone as \(\alpha: D \to \mathbb{R}^3\) with

\begin{equation*} D = \{(r,\theta) \in \mathbb{R}^2\ |\ r \in [0,R], \theta \in [0, 2 \pi] \} \end{equation*}

and

\begin{equation*} \alpha(r,\theta) = \left(r \cos(\theta), r \sin(\theta), \frac{r H}{R} \right). \end{equation*}

The tangent vectors are

\begin{equation*} \mathbf{T}_r = \left(\cos(\theta), \sin(\theta), \frac{H}{R}\right), \qquad \mathbf{T}_\theta = \left( - r \sin(\theta), r \cos(\theta), 0 \right). \end{equation*}

The cross product is

\begin{equation*} \mathbf{T}_r \times \mathbf{T}_\theta = \left( - \frac{H r}{R} \cos(\theta), - \frac{H r}{R} \sin(\theta), r \right). \end{equation*}

Its norm is

\begin{equation*} |\mathbf{T}_r \times \mathbf{T}_\theta | = \sqrt{\frac{H^2 r^2}{R^2} \cos^2(\theta) + \frac{H^2 r^2}{R^2} \sin^2(\theta) + r^2} = r \sqrt{\frac{H^2}{R^2}+1}, \end{equation*}

since \(r \in [0,R]\) and hence is positive. We then calculate the surface area:

\begin{align*} \iint_S\ dS=\amp \sqrt{\frac{H^2}{R^2}+1}\int_0^R \int_0^{2 \pi} r\ d\theta dr\\ =\amp 2 \pi \sqrt{\frac{H^2}{R^2}+1} \int_0^R r\ dr\\ =\amp \pi R^2 \sqrt{\frac{H^2}{R^2}+1} \\ =\amp \pi R \sqrt{H^2 + R^2}. \end{align*}

3.

Evaluate the unoriented surface integral

\begin{equation*} \iint_S x y \ dS, \end{equation*}

where \(S\) is the part of the plane \(x + 2 y + z = 4\) that lies in the first octant.

Solution.

We first need to parametrize the surface. We want the part of the plane that lies in the first octant, so we must have \(x \geq 0\text{,}\) \(y \geq 0\text{,}\) and \(z \geq 0\text{.}\) The equation of the plane is \(z = 4 - x - 2 y\text{.}\) Since \(z \geq 0\) and \(y \geq 0\text{,}\) we know that \(x\) cannot be more than \(4\text{.}\) So we take \(x \in [0,4]\text{.}\) Then, since \(z \geq 0\text{,}\) we know that \(4-x-2y \geq 0\text{,}\) which means that \(2 y \leq 4-x\text{,}\) that is, \(y \leq 2 - \frac{x}{2}\text{.}\) So we know that \(0 \leq y \leq 2 - \frac{x}{2}\text{.}\)

Now that we identified the region in the \((x,y)\)-plane over which the part of the plane is, we can parametrize it as \(\alpha:D \to \mathbb{R}^3\text{,}\) with

\begin{equation*} D = \{(u,v) \in \mathbb{R}^2\ | u \in [0,4], 0 \leq v \leq 2 - \frac{u}{2} \}, \end{equation*}

and

\begin{equation*} \alpha(u,v) = \left( u, v, 4 - u - 2 v \right). \end{equation*}

The tangent vectors are

\begin{equation*} \mathbf{T}_u = (1,0,-1), \qquad \mathbf{T}_v = (0,1,-2), \end{equation*}

with cross product

\begin{equation*} \mathbf{T}_u \times \mathbf{T}_v = (1,2,1), \end{equation*}

whose norm is

\begin{equation*} |\mathbf{T}_u \times \mathbf{T}_v | = \sqrt{1+2^2+1} = \sqrt{6}. \end{equation*}

We calculate the unoriented surface integral:

\begin{align*} \iint_S x y \ dS =\amp \sqrt{6} \int_0^4 \int_0^{2 - \frac{u}{2}} u v\ dv du\\ =\amp \frac{\sqrt{6}}{2} \int_0^4 u \left(2-\frac{u}{2}\right)^2\ du\\ =\amp \frac{\sqrt{6}}{2} \int_0^4 \left(4 u - 2 u^2 + \frac{u^3}{4} \right)\ du\\ =\amp \frac{\sqrt{6}}{2} \left(2 (4^2) - \frac{2}{3} 4^3 + \frac{4^4}{16} \right)\\ =\amp \frac{8\sqrt{6}}{3}. \end{align*}

4.

Consider the surface \(S = \{ y = f(x) \} \subset \mathbb{R}^3\text{,}\) where \(f: \mathbb{R} \to \mathbb{R}\) is a smooth function, and \(x \in [0,2]\text{,}\) \(z \in [0,1]\text{.}\) Show that the surface area of \(S\) is equal to the arc length of the curve \(y=f(x)\text{,}\) \(x \in [0,2]\text{,}\) in the \((x,y)\)-plane.

Solution.

We can parametrize the surface \(S\) by \(\alpha:D \to \mathbb{R}^3\) with

\begin{equation*} D = \{ (u,v) \in \mathbb{R}^2\ |\ u \in [0,2], v\in [0,1] \}, \end{equation*}

and

\begin{equation*} \alpha(u,v) = (u,f(u), v). \end{equation*}

The tangent vectors are

\begin{equation*} \mathbf{T}_u = (1,f'(u),0), \qquad \mathbf{T}_v =(0,0,1), \end{equation*}

the cross product is

\begin{equation*} \mathbf{T}_u \times \mathbf{T}_v = (f'(u),-1,0), \end{equation*}

whose norm is

\begin{equation*} |\mathbf{T}_u \times \mathbf{T}_v| = \sqrt{1+ (f'(u))^2}. \end{equation*}

The surface area is then

\begin{align*} \iint_S\ dS=\amp \int_0^2 \int_0^1 \sqrt{1+(f'(u))^2}\ dv du\\ =\amp \int_0^2 \sqrt{1+(f'(u))^2}\ du. \end{align*}

But this is precisely the formula for the arc length of the parametric curve \(\beta: [0,2] \to \mathbb{R}^2\) with \(\beta(u) = (u,f(u))\text{,}\) which is the curve \(\{y=f(x)\} \subset \mathbb{R}^2\) with \(x \in [0,2]\text{.}\)

By the way, this result is not surprising. Indeed, the surface \(S\) is basically just the curve \(y=f(x)\) extended uniformly in the \(z\)-direction. So the surface area should be the arc length of the curve times the length of the surface in the \(z\)-direction. Since \(S\) is defined with \(z \in [0,1]\text{,}\) the length in the \(z\)-direction is just \(1\text{,}\) so we the surface area should be equal to the arc length of the curve, as we obtained.

5.

In single-variable calculus, you saw that the surface area of the solid of revolution obtained by rotating the curve \(y=f(x)\text{,}\) \(x \in [a,b]\) (and \(a \geq 0\)), about the \(y\)-axis is given by the definite integral

\begin{equation*} A = 2 \pi \int_a^b x \sqrt{1+(f'(x))^2}\ dx. \end{equation*}

Show that this is consistent with our formula for the surface area of parametric surfaces.

Solution.

The surface \(S\) obtained by rotating the curve \(y=f(x)\text{,}\) \(x\in[a,b]\text{,}\) about the \(y\)-axis, can be parametrized as follows. First, we can parametrize the curve \(y=f(x)\) in the \((x,y)\)-plane as \((u,f(u),0)\text{,}\) \(u \in [a,b]\text{.}\) When we rotate the curve about the \(y\)-axis, for a fixed value of \(u\text{,}\) the point \((u,0)\) in the \((x,z)\)-plane gets rotated about the \(y\)-axis around a circle of radius \(u\text{.}\) So we should replace \((u,0)\) by \((u \cos(\theta), u \sin(\theta) )\text{.}\) This gives a parametrization for the surface \(S\) as \(\alpha: D \to \mathbb{R}^3\) with

\begin{equation*} D = \{(u,\theta) \in \mathbb{R}^2\ |\ u \in [a,b], \theta \in [0,2\pi] \} \end{equation*}

with

\begin{equation*} \alpha(u,\theta) = (u \cos(\theta), f(u), u \sin(\theta) ). \end{equation*}

The tangent vectors are

\begin{equation*} \mathbf{T}_u = (\cos(\theta), f'(u), \sin(\theta)), \qquad \mathbf{T}_\theta = (- u \sin(\theta), 0 , u \cos(\theta) ). \end{equation*}

The cross product is

\begin{equation*} \mathbf{T}_u \times \mathbf{T}_\theta = ( u f'(u) \cos(\theta), - u, u f'(u) \sin(\theta) ). \end{equation*}

Its norm is

\begin{equation*} |\mathbf{T}_u \times \mathbf{T}_\theta | = \sqrt{u^2 (f'(u))^2 \cos^2(\theta) + u^2 (f'(u))^2 \sin^2(\theta) + u^2} = u \sqrt{1 + (f'(u))^2}, \end{equation*}

since \(u \in [a,b]\text{,}\) \(a \geq 0\text{,}\) and hence \(u\) is positive. The surface area is then:

\begin{align*} \iint_S\ dS=\amp \int_a^b \int_{0}^{2 \pi} u \sqrt{1 + (f'(u))^2} d\theta du\\ =\amp 2 \pi \int_a^bu \sqrt{1 + (f'(u))^2}\ du, \end{align*}

which is the formula that you obtained in single-variable calculus!