Subdivide the curve and connect points in the subdivision by line segments. For a typical line segment, draw a right triangle with hypotenuse \(\Delta s_i\) and legs \(\Delta x_i\) and \(\Delta y_i\), and apply the Pythagorean theorem. Arc length is approximately \[\begin{array}{rcl} L \approx \sum_{i=1}^n \Delta s_i & = & \sum_{i=1}^n \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} \\ & = & \sum_{i=1}^n \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2} \Delta x_i \\ & = & \sum_{i=1}^n \sqrt{\left(\frac{\Delta x_i}{\Delta y_i}\right)^2+1} \Delta y_i. \end{array}\] Note: the subdivisions of the curve simultaneously define \(\Delta x\) and \(\Delta y\), so we can use either.

\[\begin{array}{rcl} L & = & \lim_{n\to\infty} \sum_{i=1}^n \Delta s_i \\ & = & \int_A^B ds \\ & = & \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \\ & = & \int_{y_1}^{y_2} \sqrt{\left(\frac{dx}{dy}\right)^2+1} \, dy. \end{array}\]

1. Find the length of the arc on the curve \(x^2 = 8y^3\) between \((0,0)\) and \((64,8)\). Plot the curve.

2. Find the length of the arc on the curve \(12xy - 4y^4 = 3\) between \((\frac{7}{12},1)\) and \((\frac{67}{24},2)\). Plot the curve in GeoGebra.

Consider a cone with circular base of radius \(r\) and lateral side length \(L\) (or slant length). What's the circumference of the base? \(2 \pi r\). Cut the cone along a line segment connecting the tip to the circular base and unroll it, so that the surface area of the cone is a fraction of the area of a circle of radius \(L\). What's the circumference of the circle of radius \(L\)? (Bring Ellison die circle cutouts to class. Have students fold it into quarters and see how to get a cone.) \[A_{cone} = \left( \frac{2\pi r}{2\pi L} \right) \pi L^2 = \pi r L.\]

Slice the wizard hat into a small cone of radius \(r_1\) and lateral side length \(L_1\) and a frustum radii \(r_1\) (top) and \(r_2\) (bottom) and lateral side length \(L_2\). Their combined surface area is that of a large cone of radius \(r_2\) and lateral side length \(L_1+L_2\). Thus, \(A_{\text{small cone}} + A_{\text{frustum}} = A_{\text{large cone}}\) and \[\pi r_1 L_1 + A_{\text{frustum}} = \pi r_2 (L_1 + L_2)\] implies \[A_{\text{frustum}} = \pi r_2 L_1 + \pi r_2 L_2 - \pi r_1 L_1.\]

By similar triangles, \(\frac{r_1}{L_1} = \frac{r_2}{L_1+L_2}\) and so \(r_2 L_1 = r_1(L_1+L_2)\). Thus, \[\begin{array}{rcl} A_{\text{frustum}} & = & \pi r_2 L_1 + \pi r_2 L_2 - \pi r_1 L_1 \\ & = & \pi r_1(L_1+L_2) + \pi r_2 L_2 - \pi r_1 L_1 \\ & = & \pi(r_1+r_2)L_2 \\ & = & 2\pi \frac{r_1+r_2}{2} L_2 \\ & = & 2\pi r L_2 \\ \end{array}\] where \(r = (r_1+r_2)/2\) is the average radius and \(L_2\) is the lateral side length of the frustum. Thus, the area of a frustum is the same as the area of a cylinder (with average radius)!

Idea: Subdivide the curve just like for arc length. Revolve each line segment to generate frustums. Add up the areas of the frustums. Let the number of subdivisions go to infinity. Note: \(\Delta s = L_2\). \[A = \lim_{n\to\infty} \sum_{i=1}^n 2\pi r \Delta s = \int_A^B 2\pi r \, ds\] where \(ds = \sqrt{(dx)^2+(dy)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \sqrt{\left(\frac{dx}{dy}\right)^2+1} dy\) as before.

1. Find the surface area of the paraboloid obtained by revolving \(y = 1 + 2x^2\), \(0 \leq x \leq 2\) about the \(y\)-axis.

2. A French horn has a screw-on bell in the shape of \(y = \sqrt[3]{x}\), \(1 \leq x \leq 8\) rotated about the \(y\)-axis. Find the surface area of this screw-on bell.

Revolve the unbounded region \(y = 1/x\), \(x \geq 1\) about the \(x\)-axis.

1. What is the volume of this solid of revolution?

2. What is the surface area of this solid of revolution? Hint: is \(\sqrt{1 + x^{-4}}\) bounded below?

Talk about filling Gabriel's horn with paint versus painting its surface.

Consider the astroid \(x^{2/3} + y^{2/3} = 4\).

1. Show its arc length is \(48\).

2. Show that when it is revolved about the \(x\)-axis, the surface area is \(\frac{1280\pi}{3}\).