Surface area example

Who is this for?

Quick recap of the surface area integral

• Parameterize the surface. In other words, find a vector-valued function $\vec{\textbf{v}}(\blueE{t}, \redE{s})$start bold text, v, end bold text, with, vector, on top, left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis which maps some region $T$T of the two-dimensional $\blueE{t}\redE{s}$start color #0c7f99, t, end color #0c7f99, start color #bc2612, s, end color #bc2612-plane onto your surface in three dimensions. Sometimes this parameterization will be given to you, if that's how your surface is defined. Other times the surface is defined some other way, and you have to find it yourself.

• Imagine chopping up the parameter space with horizontal and vertical lines, thus dividing your region $T$T into little rectangles. Each of these rectangles gets mapped onto a little piece of your surface which is well-approximated by a parallelogram. If your little rectangle sits at the point $(\blueE{t}, \redE{s})$left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis, and it has width $\blueE{dt}$start color #0c7f99, d, t, end color #0c7f99 and height $\redE{ds}$start color #bc2612, d, s, end color #bc2612, you can approximate its area with the following expression:

  $\begin{aligned} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right|\,\blueE{dt}\,\redE{ds} \end{aligned}$

  The smaller your initial rectangles, the more closely the corresponding piece of your surface resembles an actual flat parallelogram, and the closer this expression is to giving the true area of that piece.
• Add up the areas of these pieces with a double integral:

  $\begin{aligned} \iint_T \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right|\,\blueE{dt}\,\redE{ds} \end{aligned}$

Surface area of a torus

• Let's say the distance between the origin and the innermost part of this jelly filling is $3$3. Call this the "outer radius".
• Let's also say the distance between the innermost part of the jelly filling and the glaze itself is $1$1. Call this the "inner radius"

Step 1: Compute each partial derivative

$\begin{aligned} \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) = \end{aligned}$

$\hat{\textbf{i}} +$start bold text, i, end bold text, with, hat, on top, plus

$\hat{\textbf{j}} +$start bold text, j, end bold text, with, hat, on top, plus

$\hat{\textbf{k}}$start bold text, k, end bold text, with, hat, on top

Sprawdź

[Odpowiedź]

$\begin{aligned} \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) = \end{aligned}$

$\hat{\textbf{i}} +$start bold text, i, end bold text, with, hat, on top, plus

$\hat{\textbf{j}} +$start bold text, j, end bold text, with, hat, on top, plus

$\hat{\textbf{k}}$start bold text, k, end bold text, with, hat, on top

Sprawdź

[Odpowiedź]

Step 2: Compute the cross product

$\dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) =$start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #0c7f99, t, end color #0c7f99, end fraction, left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis, times, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #bc2612, s, end color #bc2612, end fraction, left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis, equals

$\hat{\textbf{i}} +$start bold text, i, end bold text, with, hat, on top, plus

$\hat{\textbf{j}} +$start bold text, j, end bold text, with, hat, on top, plus

$\hat{\textbf{k}}$start bold text, k, end bold text, with, hat, on top

Sprawdź

[Odpowiedź]

Step 3: Find the magnitude of this cross product

$\begin{aligned} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| = \end{aligned}$

Sprawdź

[Odpowiedź]

Step 4: Set up the appropriate double integral

Wybierz 1 odpowiedź:

Wybierz 1 odpowiedź:

• (Choice A)  

  $\begin{aligned} \int_0^{2\pi} \int_0^{6\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$

  A

  $\begin{aligned} \int_0^{2\pi} \int_0^{6\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$
• (Choice B)  

  $\begin{aligned} \int_0^{6\pi} \int_0^{2\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$

  B

  $\begin{aligned} \int_0^{6\pi} \int_0^{2\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$
• (Choice C)  

  $\begin{aligned} \int_0^{2\pi} \int_0^{2\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$

  C

  $\begin{aligned} \int_0^{2\pi} \int_0^{2\pi} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, \redE{s}) \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(\blueE{t}, \redE{s}) \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$

Sprawdź

[Odpowiedź]

Step 5: Compute the double integral

Surface area of this torus:

Sprawdź

[Odpowiedź]

Congratulations