Surface Area of Parametric Equation

Calculus Level 4

$\mathcal{P} := \begin{cases} y= \tan t \\ x = \sec t \end{cases}$

The curve $\mathcal{P}$ is defined using the above pair of parametric equations for all real values of $t$ such that $\cos t \neq 0$ .

What is the surface area of the solid of revolution obtained by revolving $\mathcal{P}$ about $x$ -axis for $0 < t < \frac{\pi}{4}$ ?

1 solution

Pranshu Gaba
Mar 27, 2016

The surface area of a solid generated by revolving a curve $(x(t), y(t))$ about the x axis for the interval $a \leq t \leq b$ is given by

$\text{Surface Area} = \int \limits _{a} ^{b} 2 \pi y \sqrt{ \left(\frac{dy}{dt}\right)^{2} + \left(\frac{dx}{dt}\right)^{2} } \, dt$