Question

The part of straight line $y=x$ between $x=1$ and $x=2$ is revolved about $X$-axis, then the curved surface area of the solid thus generated is

• A. $2\sqrt{2}\pi$
• B. $3\sqrt{2}$
• C. $3\sqrt{2}\pi$
• D. None of these

Answer 1

The correct option is C $3\sqrt{2}\pi$

Curved surface area $={\int }_a^{b}2\pi y\sqrt{\left\{{1+\left(\frac{dy}{dx}\right)}^2\right\}}dx$
Here, $a=1,b=2$ and $y=x$
Now, on differentiating $y$ with respect to $x$, we get
$\frac{dy}{dx}=1$
$\therefore$ Curved surface area $={\int }_1^{2}2\pi x\sqrt{\{1+1\}}dx$
$=2\sqrt{2}\pi {\int }_1^{2}xdx=2\sqrt{2}\pi {\left[\frac{x^2}{2}\right]}_1^2$
$=\frac{2\sqrt{2}}{2}\pi [4-1]=3\sqrt{2}\pi$ sq unit