Notes: `y = 1 + sec(x), y = 3` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

Saturday, 31 October 2015

$\displaystyle{y}={1}+{\sec{{\left({x}\right)}}},{y}={3}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

$\displaystyle{y}={1}+{\sec{{\left({x}\right)}}},{y}={3}$

Refer the image. From the graph, the curves intersects at x=-pi/3 and x=pi/3.

Using washer method,

A cross section is a washer of cross sectional area A(x) with,

Inner radius=(1+sec(x))-1=sec(x)

Outer radius=3-1=2

$\displaystyle{A}{\left({x}\right)}=\pi{\left({{2}}^{{2}}-{{\left({\sec{{\left({x}\right)}}}\right)}}^{{2}}\right)}$

$\displaystyle{A}{\left({x}\right)}=\pi{\left({4}-{{\sec}}^{{2}}{\left({x}\right)}\right)}$

Volume of the solid obtained by rotating the region bounded by the given curves about y=1 (V) is,

$\displaystyle{V}={\int_{{-\frac{\pi}{{3}}}}^{{\frac{\pi}{{3}}}}}{A}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={\int_{{-\frac{\pi}{{3}}}}^{{\frac{\pi}{{3}}}}}\pi{\left({4}-{{\sec}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={2}\pi{\int_{{0}}^{{\frac{\pi}{{3}}}}}{\left({4}-{{\sec}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={2}\pi{{\left[{4}{x}-{\tan{{\left({x}\right)}}}\right]}_{{0}}^{{\frac{\pi}{{3}}}}}$

$\displaystyle{V}={2}\pi{\left({\left({4}\cdot\frac{\pi}{{3}}-{\tan{{\left(\frac{\pi}{{3}}\right)}}}\right)}-{\left({4}\cdot{0}-{\tan{{\left({0}\right)}}}\right)}\right)}$

$\displaystyle{V}={2}\pi{\left({4}\cdot\frac{\pi}{{3}}-\sqrt{{{3}}}-{0}\right)}$

$\displaystyle{V}={2}\pi{\left(\frac{{{4}\pi}}{{3}}-\sqrt{{{3}}}\right)}$