Surface Area for Parametric Equations

10.2 Surface Area for Parametric Equations

Introduction

This section explores how to find the surface area generated by rotating a curve defined by parametric equations about the x-axis. This is a key application of integration in calculus, particularly when dealing with curves that are not easily described by explicit functions.

Surface Area of a Solid of Revolution (Parametric Form)

• Definition: The surface area generated by rotating a parametric curve about the x-axis can be found using the following formula:

$S = \int_{a}^{b} 2\pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

• Where:

  • $x = f(t)$ and $y = g(t)$ are the parametric equations for the curve.

  • $a \leq t \leq b$ defines the interval for the parameter $t$.

• Explanation: The formula calculates the surface area by summing up the areas of infinitesimal bands formed as the curve is rotated about the x-axis.

Example: Surface Area of a Rotated Semicircle

• Given Parametric Equations:

  • $x = 5 \cos(t)$

  • $y = 5 \sin(t)$

  • $0 \leq t \leq \pi$

• Interpretation: These equations describe a semicircle of radius 5 centered at the origin, traced from the rightmost point to the leftmost point as $t$ goes from $0$to$\pi$.

• Application: Rotating this semicircle about the x-axis generates a sphere.

• Surface Area Calculation:

  • Plug the parametric equations into the surface area formula:

  $S = \int_{0}^{\pi} 2\pi (5 \sin t) \sqrt{\left(\frac{d}{dt}(5 \cos t)\right)^2 + \left(\frac{d}{dt}(5 \sin t)\right)^2} \, dt$

  • Compute derivatives:

  • $\frac{dx}{dt} = -5 \sin t$

  • $\frac{dy}{dt} = 5 \cos t$

  • So,

  $\sqrt{(-5 \sin t)^2 + (5 \cos t)^2} = \sqrt{25 \sin^2 t + 25 \cos^2 t} = \sqrt{25(\sin^2 t + \cos^2 t)} = 5$

  • Therefore, the integral simplifies to:

  $S = \int_{0}^{\pi} 2\pi (5 \sin t) (5) \, dt = 50\pi \int_{0}^{\pi} \sin t \, dt$

  • Evaluate the integral:

  $\int_{0}^{\pi} \sin t \, dt = [-\cos t]_{0}^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2$

  • Final surface area:

  $S = 50\pi \times 2 = 100\pi$

• Conclusion: The surface area of the sphere generated by rotating the semicircle is $100\pi$.

Special Case: Surface Area of a Sphere

• The general formula for the surface area of a sphere of radius $r$ is:

$A = 4\pi r^2$

• For $r = 5$, $A = 4\pi (5)^2 = 100\pi$, which matches the result above.

Summary Table: Surface Area by Parametric Rotation

Curve Type	Parametric Equations	Interval	Surface Area Formula
General Curve	$x = f(t)$, $y = g(t)$	$a \leq t \leq b$	$S = \int_{a}^{b} 2\pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
Semicircle (radius $r$)	$x = r \cos t$, $y = r \sin t$	$0 \leq t \leq \pi$	$S = 4\pi r^2$

Additional info: The "Desmos Graph: Surface of Area Sphere" reference suggests using graphing technology to visualize the surface area generated by rotation, which can aid in understanding the geometric interpretation of the integral.