2 Attachment(s)
Calculate Volume using the Method of Cylindrical Shells - Urgent
Attachment 25095
Here are some things I have done. I already found the volume since the other part of the question said to use the method of discs/washers. It is 54.0633 cubic units (correct answer).
I found by trial and error that (since this is an online assignment) that a= 9/sqrt(82) and c =9 .
What do I do? Why should there be two integrals, I don't understand.
I have tried the following.
y = 9^x becomes x = lny/ln9
Formula
h = lny/ln9
r = y
So I got integral from 9/sqrt(82) to 9 of
Attachment 25096 dy
But that's just one integral and perhaps even incorrect. Please show me what to do.
Thanks
Re: Calculate Volume using the Method of Cylindrical Shells - Urgent
the sketch of the graph is misleading ...
note that 
, so there is a very small region between the curve 
and 
from 
to 
.
Re: Calculate Volume using the Method of Cylindrical Shells - Urgent
To evaluate the volume of your region, you need to do
![\displaystyle \begin{align*} \int_1^9{ \pi \cdot 1^2 \, dy } - \int_1^9{ \pi \cdot \left( \log_9{y} \right)^2 \,dy } + \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot 1^2 \, dy } - \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot \left[ \sqrt{ \left( \frac{y}{9} \right)^{ -2 } - 81 } \right]^2 \,dy } &= \pi \int_1^9{ 1 \, dy } - \frac{\pi}{\left( \ln{9} \right)^2} \int_1^9{ \left( \ln{y} \right)^2 \,dy } + \pi \int_{ \frac{9}{\sqrt{82}} }^1{ 1\, dy } - 81\pi \int_{\frac{9}{\sqrt{82}}}^1{ y^{-2} - 1 \, dy } \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \int_1^9{ \pi \cdot 1^2 \, dy } - \int_1^9{ \pi \cdot \left( \log_9{y} \right)^2 \,dy } + \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot 1^2 \, dy } - \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot \left[ \sqrt{ \left( \frac{y}{9} \right)^{ -2 } - 81 } \right]^2 \,dy } &= \pi \int_1^9{ 1 \, dy } - \frac{\pi}{\left( \ln{9} \right)^2} \int_1^9{ \left( \ln{y} \right)^2 \,dy } + \pi \int_{ \frac{9}{\sqrt{82}} }^1{ 1\, dy } - 81\pi \int_{\frac{9}{\sqrt{82}}}^1{ y^{-2} - 1 \, dy } \end{align*})
The only one of those integrals which may cause you any trouble is the second, so do
^2 \, dy} &= \int{ \frac{y\left( \ln{y} \right)^2}{y} \, dy } \\ &= \int{ u^2 \, e^u \, du } \textrm{ after making the substitution } u = \ln{y} \end{align*})
which you would then evaluate using integration by parts twice. See how you go :)
Re: Calculate Volume using the Method of Cylindrical Shells - Urgent
Is this the method of cylindrical shells? I am obliged to use cylindrical shells. Can you use r (radius), h(height)? And why will my answer be in the form of a sum of two integrals (see the question -top of image)?
Re: Calculate Volume using the Method of Cylindrical Shells - Urgent
using the method of cylindrical shells ...
 \, dy + 2\pi \int_1^9 y \left(1 - \frac{\ln{y}}{\ln{9}}\right) \, dy)
Re: Calculate Volume using the Method of Cylindrical Shells - Urgent
