Browse Questions

The volume, When the curve $y=\sqrt{3+x^{2}}$ from $x=0$ to $x=4$ is rotated about $x$-axis is

$\begin{array}{1 1}(1)100\pi&(2)\frac{100}{9}\pi\\(3)\frac{100}{3}\pi&(4)\frac{100}{3}\end{array}$

Required volume $V= \int \limits_a^b \pi y^2 dx$
$\qquad= \pi \int \limits_0^4 (3+x^2) dx$
$\qquad= \pi [3x+\large\frac{x^2}{3} ]^4_0$
$\qquad= \pi [3(4-0) +\large\frac{1}{3}$$(4^3 -0)] \qquad= \pi [12 +\large\frac{64}{3}] \qquad = \pi [ \large\frac{36+64}{3}]=\frac{100}{3}$$\pi$
Hence 3 is the correct answer.