Shell Method Solution 5

Processing math: 100%

Solution a.): Here is a carefully labeled sketch of the region with a shell marked on the

$x$ -axis at

$x$ . The shell has radius

$r$ , measured from the

$y$ -axis, and height

$h$ , taken parallel to the

$y$ -axis at

$x$ . It is IMPORTANT to mark ALL of

$x$ ,

$r$ , and

$h$ in the sketch of the region !!!

tex2html_wrap_inline125

Thus the total volume of this Solid of Revolution is

$Volume = 2 \pi \int_{0}^{\pi/3} (radius)(height) \ dx = 2 \pi \int_{0}^{\pi/3} rh \ dx$

$= 2 \pi \int_{0}^{\pi/3} (x)(\sqrt{3}- \tan x) \ dx$

Solution b.): IMPORTANT CHANGE: Because we are revolving the region about the

$x$ -axis, we must mark a shell on the

$y$ -axis at

$y$ !!! The shell has radius

$r$ , measured from the

$x$ -axis, and height

$h$ , taken parallel to the

$x$ -axis at

$y$ . It is IMPORTANT to mark ALL of

$y$ ,

$r$ , and

$h$ in the sketch of the region !!!

tex2html_wrap_inline125

Thus the total volume of this Solid of Revolution is

$Volume = 2 \pi \int_{0}^{\sqrt{3}} (radius)(height) \ dy = 2 \pi \int_{0}^{\sqrt{3}} rh \ dy$

$= 2 \pi \int_{0}^{\sqrt{3}} (y)(\arctan y) \ dy$

Click HERE to return to the list of problems.