Shell Method Solution 3

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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 !!!

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Thus the total volume of this Solid of Revolution is

$Volume = 2 \pi \int_{1}^{e^3} (radius)(height) \ dx = 2 \pi \int_{1}^{e^3} rh \ dx$

$= 2 \pi \int_{1}^{e^3} (x)(3- \ln x) \ dx$

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

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

$= 2 \pi \int_{0}^{3} (y)(e^y-1) \ dy$

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