Question

The area bounded by the curve $y={\tan }^{2}x$ and lines $x=0,x=\frac{\pi }{4}$ revolved about x-axis then volume of solid so generated is:

• A. $(\frac{{\pi }^2}{4}-\pi ){\left(unit\right)}^3$
• B. $(\pi -\frac{{\pi }^2}{4}){\left(unit\right)}^3$
• C. $(\frac{{\pi }^3}{4}-\pi ){\left(unit\right)}^3$
• D. $(\pi -\frac{{\pi }^3}{3}){\left(unit\right)}^3$

Answer 1

Answer: (B)

$(\pi -\frac{{\pi }^2}{4}){\left(unit\right)}^3$

Volume $=\pi {\int }_0^{\frac{\pi }{4}}{\tan }^{2}xdx$

$=\pi {\int }_0^{\frac{\pi }{4}}[{\sec }^{2}x-1]dx$

$=\pi [\tan \frac{\pi }{4}-\tan 0-\frac{\pi }{4}]$

$=\pi \times [1-\frac{\pi }{4}]$

$=\pi -\frac{(\pi {)}^2}{4}$