Question

A surface of revolution is obtained by rotating a curve \(z=f(x), y=0\) in the \(x z\)-plane about the \(z\)-axis. a) Show that this surface has the equation \(z=f(r)\) in cylindrical coordinates. b) Show that the area of the surface is $$ S=\int_{0}^{2 \pi} \int_{a}^{b} \sqrt{1+f^{\prime}(r)^{2}} r d r d \theta=2 \pi \int_{a}^{b} \sqrt{1+f^{\prime}(r)^{2}} r d r $$

Short answer

Question: Show that the equation of the surface of revolution in cylindrical coordinates is \(z=f(r)\), and derive the surface area formula. Answer: The surface of revolution has the equation \(z = f(r)\) in cylindrical coordinates, and the surface area formula is given by \(S = 2\pi \int_{a}^{b} \sqrt{1+f'^2(r)} r dr\).

Step-by-step solution

Part (a): Convert to cylindrical coordinates

First, let's recall the equations to convert from cartesian \((x, y, z)\) to cylindrical coordinates \((r, \theta, z)\): $$ x = r\cos{\theta}, \qquad y = r\sin{\theta}, \qquad z = z. $$ Given that the curve is \(z=f(x)\), we replace \(x\) in the equation with \(r\cos{\theta}\), obtaining $$ z = f(r\cos{\theta}). $$ Since we are told that \(y=0\), we have \(\sin{\theta}=0\). This implies either \(\theta=0\) or \(\theta=\pi\). However, rotating around the \(z\)-axis involves all values of \(\theta\), so the function \(f\) must depend only on \(r\) rather than \(r\cos{\theta}\), which gives $$ z = f(r). $$ Thus, we have shown that the surface of revolution has the equation \(z=f(r)\) in cylindrical coordinates.

Part (b): Derive the surface area formula

To derive the surface area, we first need to find the element of the surface area, \(dS\), for a tiny patch in cylindrical coordinates. Consider moving vertically along the curve \(z = f(r)\) by an increment \(r+dr\). The change in \(z\) is \(dz=f'(r)dr\). We also move horizontally along the azimuthal direction with an increment of \(d\theta\). Therefore, the small horizontal distance is \(rd\theta\). We can now find the change in \(S\). Using the Pythagorean theorem, the infinitesimal area \(dS\) is given by: $$ dS = \sqrt{(rd\theta)^2 + (f'(r)dr)^2} = \sqrt{r^2d\theta^2 + f'^2(r) dr^2}. $$ Now, we integrate over the range \(0\) to \(2\pi d\theta\) and \(a\) to \(b dr\) to find the total surface area: $$ S = \int_{0}^{2\pi} \int_{a}^{b} \sqrt{1+f'^2(r)}r dr d\theta. $$ To solve the integral, first note that the integrand does not depend on \(\theta\). Therefore, the \(\theta\) integral is just the length of the interval, which is \(2\pi\). Thus, the surface area is $$ S = 2\pi \int_{a}^{b} \sqrt{1+f'^2(r)} r dr. $$ We have shown that the area of the surface is given by the provided integral.

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are immensely helpful in tackling problems with symmetry around an axis, like surfaces of revolution. Think of cylindrical coordinates as an extension of polar coordinates with an extra component for height. Instead of describing a point's position with x and y coordinates, we use a distance from the origin, an angle around the origin, and a height.

• Radius (r): the distance from the point to the z-axis.
• Angle (θ): the counterclockwise angle in the x-y plane, measured from the positive x-axis.
• Height (z): the same as in Cartesian coordinates—how high or low the point is.

By expressing a function in cylindrical coordinates, we can simplify the equation when dealing with rotationally symmetric shapes. As seen in our exercise, converting the function into cylindrical coordinates allowed us to express a surface of revolution with the much simpler form of z=f(r), showing dependence only on the radius from the z-axis, which is crucial for finding the area of such surfaces.

Surface Area Integration

Surface area integration is a powerful application of calculus where we sum up infinitesimally small patches to find the total area of a curved surface.
Consider draping a fabric over a lampshade; the fabric must cover every part of the lampshade's surface. Calculating this fabric's exact size—without stretching—requires finding the surface area of the lampshade. Integration allows us to do just that!

Infinitesimal Area Patches (dS)

For surfaces of revolution, our 'fabric' is cut into tiny rings, each with a radius of r and thickness dr. If you were to cut one of these rings and lay it flat, you'd see it's a rectangle with length equal to the circumference of the ring, or 2πr, and width dr. However, because the surface is curved, we need to adjust this width by taking into account how steeply the surface curves, which is where f'(r) comes into play.
Integrating these adjusted infinitesimal patches from the exercise solution over the whole surface gives us the formula for the surface area:

Calculus

Calculus, the mathematical study of continuous change, is the toolbox for solving a wide range of problems in science and engineering. In our case, we're using integral calculus—a method of adding up a series of infinitesimally small quantities to find the whole.
The process of integration allows us to consider the tiny incremental changes in radius (dr) and angle (dθ), and how these relate to increments in area (dS). This precise approach leads us to the solution to part (b) of our exercise. Through integration, we have determined that the surface area of the shape can be computed with a relatively simple equation that beautifully ties together the radius and the derivative of the function.
By moving from the specific (dS, or the area of a tiny piece of the surface) to the general (S, or the total surface area), calculus gives us the power to make sense of seemingly complex and curved shapes with elegance and precision.