Basic Mathematics20775 marksNumerical: surface area of revolutionSurface area of revolution
Find the area of the surface generated by revolving the curve y = 2sqrtx, 1 leq x leq 2, about the x-axis.
Find the area of the surface generated by revolving the curve $y = 2\sqrt{x}$, $1 \leq x \leq 2$, about the x-axis. [5]
- Curve: $y = 2\sqrt{x}$ - Interval: $1 \le x \le 2$ - Axis of revolution: x-axis $$S = 2\pi \inta^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx$$ $$y = 2x^{1/2} \implies \frac{dy}{dx} = x^{-1/2} = \frac{1}{\sqrt{x}}$$ $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{x}$$ $$1 + \frac{1}{x} = \frac{x...