Integration By Parts: Volume - help

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Homework Help Overview

The problem involves finding the volume generated by rotating a region bounded by the curves \(y = e^{1.6x}\), \(y = e^{-1.6x}\), and the line \(x = 0.6\) about the y-axis using the method of cylindrical shells.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of the integral for volume using cylindrical shells, with some attempting to clarify the correct formulation of the integral. There are questions about the antiderivative and the correctness of the initial integral setup.

Discussion Status

Some participants have provided guidance on correcting the integral setup and suggested using integration by parts. There is an acknowledgment of differing interpretations regarding the region to be rotated, leading to a productive discussion about the boundaries and the curves involved.

Contextual Notes

Participants are navigating assumptions about the symmetry of the curves and the implications for setting up the volume integral. There is a recognition of the need to clarify the region of integration based on the problem statement.

Slimsta
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Homework Statement


Use the method of cylindrical shells to find the volume generated by rotating the region R bounded by the curves y=e1.6 x, y=e−1.6 x and x=0.6 about the y-axis.


Homework Equations


V=[tex]$\displaystyle \Large \int _a^c 2pix (yt - yb) dx$[/tex]

The Attempt at a Solution


[tex]$\displaystyle \Large 2pi * \int _0^{0.6} x(e^{1.6} - e^{-1.6})dx$[/tex]

is this right?
if yes, how do i find the antiderivative of it? i got an answer but its wrong for some reason..
 
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You're close, but you left off the x parts in your exponential functions. This is what you want:
[tex]2\pi * \int _0^{0.6} x(e^{1.6x} - e^{-1.6x})dx[/tex]

I would split these into two integrals and use integration by parts on each one.
 
Your integral is wrong.
Watch and learn :smile:

http://img28.imageshack.us/img28/640/volumeintegration.png
Notice how [itex]y=e^{1.6x}[/itex] and [itex]y=e^{-1.6x}[/itex] are symmetrical about the y-axis. This means we only need to deal with one side of the axis. Let's take the positive side.

http://img163.imageshack.us/img163/613/cylinder.png
We want to find the volume of the grey glittery bit.

[tex]r=x[/tex]

[tex]R=r+\delta x=x+\delta x[/tex]

[tex]h=y=e^{-1.6x}[/tex]

Now the volume is
[tex]\delta V=\pi \left(R^2-r^2\right)h[/tex]

and substituting into this formula and simplifying we have
[tex]\delta V=2\pi x e^{-1.6x} \delta x[/tex]

(we ignore the [itex](\delta x)^2[/itex] since this is a negligible amount)

So you can now set up your integral accordingly and use integration by parts to solve it. Note that we could have also taken the other side of the x-axis so instead we have [itex]h=e^{1.6x}[/itex] and the limits in our integral will be from -0.6 to 0 instead of 0 to 0.6.
 
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Mentallic,
I disagree. Slimsta's problem statement says that the region to be rotated is bounded by the two exponential curves and the line x = 0.6. So the region R is above your green-shaded region, and below the red curve, and bounded on the right by the vertical line.
 
Oh my! Yes you're right, I was completely off.

Now it makes sense as to why the question would bother adding in that other exponential curve :blushing:
 

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