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Section 3.6 Surface Area

Subsection 3.6.1 Calculation

Activity 60.

The purpose of this activity is to illustrate the surface area formula.
Continue using the surface \(S(t,u)=(e^t,e^u,tu)\) and the tangent plane \(P(t,u)\) from Activity 58.
(a)
Calculate the cross product \(\frac{\partial S}{\partial t}\) and \(\frac{\partial S}{\partial u}\text{.}\)
(b)
Calculate the norm of this cross product.
(c)
Evaluate \(P(t,u)\) at the points \((0,0)\text{,}\) \((1,0)\text{,}\) \((1,1)\text{,}\) and \((0,1)\text{.}\)
(d)
Calculate the area of the rectangle formed by these four points.
(e)
Compare these two scalar results.
(f)
How could the total surface area of this surface for \(t \in [0,10]\) and \(u \in [0,10]\) be calculated?

Exercises Exercises

For each of the following surfaces. Graph the surface using technology, then setup the integral for its surface area. Finally use technology to calculate the result. Note each of these is expressed in Cartesian coordinates.
1.
\(f(x,y)=x^2+y^2\) for \(x \in [-1,1]\) and \(y \in [-1,1].\)
2.
\(S(\theta,z)=(\cos\theta,\sin\theta,z)\) for \(\theta \in [0,2\pi]\) and \(z \in [0,1].\)
3.
\(S(\theta,z)=(z\cos\theta,z\sin\theta,z)\) for \(\theta \in [0,2\pi]\) and \(z \in [0,1].\)
4.
\(S(\theta,\phi)=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)\) for \(\theta \in [0,2\pi]\) and \(\phi \in [0,\pi].\)

Subsection 3.6.2 Surface Area in Cylindrical Coordinates

The surface area formula was developed using rectangles (width and height) which ties it to Cartesian coordinates. If we are going to calculate the surface area of solids expressed in other coordinate systems, we must first convert the coordinates.

Example 3.6.2.

Calculate the surface area of the solid represented by \(r=1-(z-1)^2=2z-z^2\) for \(z \in [0,2]\text{.}\)
First we express the surface in parametric form in Cartesian using the conversion developed in Subsection 3.1.2.
\begin{align*} x = & r\cos\theta \\ = & (2z-z^2)\cos\theta.\\ y = & r\sin\theta \\ = & (2z-z^2)\sin\theta.\\ z = & z. \end{align*}
Thus this surface is given by
\begin{equation*} S(z,\theta)=() (2z-z^2)\cos\theta, (2z-z^2)\sin\theta, z )\text{.} \end{equation*}
Now we can use the formula.
\begin{align*} \frac{\partial S}{\partial z} = & \langle (2-2z)\cos\theta, (2-2z)\sin\theta, 1 \rangle.\\ \frac{\partial S}{\partial \theta} = & \langle -(2z-z^2)\sin\theta, (2z-z^2)\cos\theta, 0 \rangle.\\ \frac{\partial S}{\partial z} \times \frac{\partial S}{\partial \theta} = & \langle -(2z-z^2)\cos\theta, -(2z-z^2)\sin\theta, 2z^3 - 6z^2 + 4z \rangle \end{align*}
Thus the surface area is
\begin{equation*} \int_0^{2\pi} \int_0^2 \left\| \frac{\partial S}{\partial z} \times \frac{\partial S}{\partial \theta} \right\| \; dz \; d\theta \approx 10.97. \end{equation*}

Subsection 3.6.3 Scalar Surface Integrals

Section 3.2 presented a method for calculating the scalar integral along a parameterized curve. Such an integral can be interpreted as the total affect of a function along a curve. This idea can be extended to calculating the scalar integral over a parameterized surface. Such an integral can be interpreted as the total affect of a function over a surface.

Activity 61.

(a)
Write the formula for the scalar line integral for a function \(f\) along a curve \(C(t)\text{.}\)
(b)
Explain the role of \(f(C(t))\) in this integral.
(c)
Explain the role of \(\sqrt{C^\prime(t)}\) in this integral.
(d)
If the temperature in an oven is uniformly \(375^\circ\) F at each point, how much heat is absorbed by the curve \(C(t)=(\cos t, \sin t,1)\text{?}\)
(e)
Based on the pattern of the scalar line integral write the formula for the scalar surface integral for a function \(f\) over a surface \(S(t,u)\text{.}\)
(f)
Explain the role of \(f(S(t,u))\) in this integral.
(g)
Explain the role of \(\left| \frac{\partial S}{\partial t} \times \frac{\partial S}{\partial u} \right| \) in this integral.
(h)
If the temperature in an oven is uniformly \(375^\circ\) F at each point, how much heat is absorbed by the surface \(\vec{s}(t,s)=(2t+3s+1,t+s+1,t-s+1)\) for \(t \in [0,1]\text{,}\) \(s \in [0,1]\text{?}\)
For a smooth surface \(S(t,u)\) the scalar surface integral can be calculated as
\begin{equation*} \dint f(S(t,u))\left\| \frac{\partial S}{\partial t} \times \frac{\partial S}{\partial u} \right\| \; dA. \end{equation*}