![[Picture]](digInSurfaceArea.online-fb3d79fcb7f557c9162e4b214e9f3ed3.svg)
\[ \text {Area} = \answer {.02\sqrt {3}+.02\sqrt {18}+.02\sqrt {42}} \]
Use the fact that
\[ \d S = \sqrt {1+F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d x \d y \]
We compute surface area with double integrals.
In the past, we’ve used definite integrals to compute the arc length of curves. The natural extension of the concept of arc length over an interval is surface area over a region. Consider the surface \(z=F(x,y)\) over some region in the \((x,y)\)-plane. To compute the surface area, first consider the surface area of a “patch” of the surface, determined by \(\vec {u}\) and \(\vec {v}\) below:
![[Picture]](digInSurfaceArea.online-ea44923be913c9c39664126dab3e2342.svg)
In essence, we zoom in on this portion of the surface to the extent that the tangent plane approximates the function so well that in this figure, it is virtually indistinguishable from the surface itself. Therefore we can approximate the surface area \(\d S\) of a “patch” of this region of the surface with the area of the parallelogram spanned by \(\vec {u}\) and \(\vec {v}\). Here
\begin{align*} \vec {u} &= \vector {\d x,0,\pp [F]{x}\d x}\\ \vec {v} &= \vector {0,\d y,\pp [F]{y}\d y}, \end{align*}
hence
\begin{align*} \d S &= |\vec u\cross \vec v|\\ &= \left |\vector {\d x,0,\pp [F]{x}\d x}\cross \vector {0,\d y,\pp [F]{y}\d y}\right |\\ &=\sqrt {1+F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d x\d y. \end{align*}
Summing these “patches” together leads to a double integral.
Let \(z=F(x,y)\) where \(\pp [F]{x}\) and \(\pp [F]{y}\) are continuous over a closed, bounded region \(R\). The surface area \(S\)
over \(R\) is
\begin{align*} S &= \iint _R \d S\\ &=\iint _R \sqrt {1+F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d A. \end{align*}
A table of gradient vectors for a function \(F:\R ^2\to \R \) is given below:
Let \(R\) be the
shaded region above. Estimate the surface area of \(F\) over \(R\).
In our discussion above, we are in fact defining the concept of the “area of a surface.”
While we already have a notion of the area of a region in the plane, we did not yet
have a solid grasp of what “the area of a surface in space” means. Double integrals
make this notion of surface area precise.
Let’s train ourselves to use our new tools by computing the surface areas of known surfaces. We start with a triangle.
Let \(F(x,y) = 4-x-2y\), and let \(R\) be the region in the \((x,y)\)-plane ![[Picture]](digInSurfaceArea.online-73b785b2497587e3bc06893cc9b25ea8.svg)
Find the surface area of \(F\) over \(R\).
\[ R = \{(x,y):0\le x\le 4\text { and } 0\le y\le 2-x/2\} \]
We start by noting that
\begin{align*} \pp [F]{x} = \answer [given]{-1},\\ \pp [F]{y} = \answer [given]{-2}. \end{align*}
Therefore, from our definition above,
\begin{align*} S &= \iint _R\d S \\ &= \int _0^{\answer [given]{4}}\int _0^{\answer [given]{2-x/2}} \sqrt {1+(-1)^2+(-2)^2}\d y\d x\\ &= \int _0^{\answer [given]{4}} \answer [given]{\sqrt {6}\left (2-\frac {x}{2}\right )}\d x\\ &= \answer [given]{4\sqrt {6}}. \end{align*}
Because the surface is a triangle, we can figure out the area using the cross product. Setting
\begin{align*} \vec {v} = \vector {-4,0,\answer [given]{4}},\\ \vec {w} = \vector {0,-2,\answer [given]{4}}, \end{align*}
we see that
\[ \vec {v} \cross \vec {w} = \vector {\answer [given]{8},\answer [given]{16},\answer [given]{8}}, \]
and so \[ |\vec {v}\cross \vec {w}| = \answer [given]{8\sqrt {6}}. \]
Since this is the area of the parallelogram spanned by \(\vec {v}\) and \(\vec {w}\), we
divide by \(\answer [given]{2}\) to confirm the area of the triangle is \(4\sqrt {6}\). It’s common knowledge that the surface area of a sphere of radius \(a\) is \(4\pi a^2\). While there are several ways to confirm this formula, we will use a double integral. Our computation will involves using our formula for surface area, polar coordinates, and improper integrals!
Find the surface area of the sphere with radius \(a\) centered at the origin.
To start, we
will compute the surface area of
\[ F(x,y) =\sqrt {a^2-x^2-y^2} \]
over the region \[ R = \{(x,y): x^2 + y^2 \le a^2\}. \]
We start by computing partial
derivatives and find \begin{align*} \pp [F]{x} &= \answer [given]{\frac {-x}{\sqrt {a^2-x^2-y^2}}} \\ \pp [F]{y} &= \answer [given]{\frac {-y}{\sqrt {a^2-x^2-y^2}}}. \end{align*}
Since our function \(F\) only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area:
\begin{align*} S &= 2\iint _R \d S \\ S &= 2\iint _R \sqrt {1+ F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d A \\ &= 2\iint _R \sqrt {1+ \frac {x^2+y^2}{a^2-x^2-y^2}}\d A. \end{align*}
Since the region \(R\) that we are integrating over is the circle, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions
\begin{align*} x&=\answer [given]{r\cos (\theta )},\\ y&=\answer [given]{r\sin (\theta )},\\ \d A &= r\d r\d \theta , \end{align*}
we write
\begin{align*} S &= 2\iint _R \sqrt {1+ \frac {x^2+y^2}{a^2-x^2-y^2}}\d A\\ &=2\int _0^{2\pi }\int _0^a \sqrt {1+\frac {r^2\cos ^2\theta +r^2\sin ^2\theta }{a^2-r^2\cos ^2\theta -r^2\sin ^2\theta }} r\d r\d \theta \end{align*}
Simplifying, we now write:
\[ S = 2\int _0^{2\pi }\int _0^a \sqrt {\frac {a^2}{a^2-r^2}} r \d r \d \theta \]
We evaluate this integral by making the substitution
\begin{align*} g &= a^2 -r^2,\\ \d g &= \answer [given]{-2r}\d r,\\ \d r &= \frac {\d g}{\answer [given]{-2 r}}, \end{align*}
and paying close attention to the limits of integration, we now have
\[ S = 2\int _0^{2\pi }\int _{a^2}^0\frac {a g^{-1/2}}{-2} \d g \d \theta \]
However, since
the integrand of the inner integral is not defined at \(g= 0\), the inner integral is in fact an
improper integral. Let’s carefully evaluate the inner integral. Write with me:
\begin{align*} \int _{a^2}^0\frac {a g^{-1/2}}{-2} \d g &= \lim _{b\to 0}\int _{a^2}^b\frac {a g^{-1/2}}{-2} \d g\\ &= \lim _{b\to 0}\eval {\answer [given]{-a \sqrt {g}}}_{a^2}^b\\ &= \lim _{b\to 0}(-a \sqrt {b}+ a \sqrt {a^2})\\ &= \answer [given]{a^2}. \end{align*}
So we may now write
\begin{align*} S &= 2\int _0^{2\pi }\int _{a^2}^0\frac {a g^{-1/2}}{-2} \d g \d \theta \\ &= 2\int _0^{2\pi } \answer [given]{a^2} \d \theta \\ &= \answer [given]{4\pi a^2}. \end{align*}
Thus confirming our previous formula.
Let’s find the surface area of a general region now.
Find the area of the surface \(F(x,y) = x^2-3y+3\) over the region ![[Picture]](digInSurfaceArea.online-9c07f212beb6e8edc69f0d46eabda2b9.svg)
\[ R = \{(x,y):-x\le y\le x\text { and } 0\le x\le 4\} \]
To start, compute the partial derivatives:
\begin{align*} \pp [F]{x} &= \answer [given]{2x}\\ \pp [F]{y} &= \answer [given]{-3} \end{align*}
Thus the surface area is described by the double integral
\[ \iint _R \sqrt {10+4x^2}\d A. \]
As with integrals describing
arc length, double integrals describing surface area are in general hard to evaluate
directly because of the square-root. This particular integral can be easily evaluated,
though, with judicious choice of our order of integration. Integrating with order \(\d x\d y\)
requires us to evaluate \[ \int \sqrt {10+4x^2}\d x. \]
This is not too easy. The computation involves integration by
parts and \(\sinh ^{-1}(x)\).
However, integrating in the order \(\d y\d x\) has as its first integral
\[ \int \sqrt {10+4x^2}\d y, \]
and this is easy to evaluate,
\[ \int \sqrt {10+4x^2}\d y = \answer [given]{y\sqrt {10+4x^2}}+C. \]
So we proceed with the order \(\d y\d x\). In this case, the limits of integration are already
given in the statement of the problem. Write with me: \begin{align*} \iint _R\sqrt {10+4x^2}\d A &= \int _0^4\int _{-x}^x\sqrt {10+4x^2}\d y \d x\\ &= \int _0^4\eval {y\sqrt {10+4x^2}}_{-x}^x \d x\\ &=\int _0^4\big (2x\sqrt {10+4x^2}\big )\d x \end{align*}
Setting, \(g=10+4x^2\) you can use substitution to find:
\begin{align*} \int _0^4\big (2x\sqrt {10+4x^2}\big )\d x &= \eval {\answer [given]{\frac {1}{6}\big (10+4x^2\big )^{3/2}}}_0^4 \\ &= \frac {1}{3}\big (37\sqrt {74}-5\sqrt {10}\big ) \end{align*}
Note that the surface has a much greater area than the region in the \((x,y)\)-plane. This is because the \(z\)-values of the surface change dramatically over \(R\).
In practice, technology helps greatly in the evaluation of such integrals. High powered computer algebra systems can compute integrals that are difficult, or at least time consuming, by hand, and can at the least produce very accurate approximations with numerical methods. In general, just knowing how to set up the proper integrals brings one very close to being able to compute the needed value. Most of the work is actually done in just describing the region \(R\) in terms of polar or rectangular coordinates. Once this is done, technology can usually provide a good answer.