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Finding a volume bounded by a region

  • Thread starter l46kok
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  • #1
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Homework Statement


The region bounded by [itex]y=e^-x^2[/itex], y=0, x=0 and x=1 is revolved around the y-axis. Find the volume.


Homework Equations



[itex]V = \int_{a}^{b} R^2 dy[/itex]

The Attempt at a Solution



We must first express x in terms of y. So we get

[itex]\ln y = -x^2[/itex]

[itex]x^2 = -\ln y[/itex]

We substitute this to the above volume equation and we get

[itex]V = \int_{0}^{1} -\ln y[/itex]

And without even doing the integration, I know for a fact that this cannot work out since if you integrate a natural log, you get a natural log back and substituting 0 into a natural log gets you an undefined answer.

I think I did something wrong or set up the problem wrong. Any advice would be appreciated!

Oh and if you know latex, could you tell me how to express y=e^(-x^2)?
 

Answers and Replies

  • #2
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Homework Statement


The region bounded by [itex]y=e^-x^2[/itex], y=0, x=0 and x=1 is revolved around the y-axis. Find the volume.


Homework Equations



[itex]V = \int_{a}^{b} R^2 dy[/itex]
This "relevant" equation isn't relevant.

The Attempt at a Solution



We must first express x in terms of y. So we get

[itex]\ln y = -x^2[/itex]

[itex]x^2 = -\ln y[/itex]

We substitute this to the above volume equation and we get

[itex]V = \int_{0}^{1} -\ln y[/itex]

And without even doing the integration, I know for a fact that this cannot work out since if you integrate a natural log, you get a natural log back and substituting 0 into a natural log gets you an undefined answer.

I think I did something wrong or set up the problem wrong.
Yes. The easiest way to do this problem is to use cylindrical shells.

The relevant equation is
[tex]V = \int_a^b 2\pi~radius * height~dx[/tex]

Here, radius = x, and height = y = e-x2
Any advice would be appreciated!

Oh and if you know latex, could you tell me how to express y=e^(-x^2)?
Put the exponent in braces:{-x^2}.
[tex]y = e^{-x^2}[/tex]
 
  • #3
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Just so that I understand, why is it that the cylindrical shell method must be applied here? I was taught that when finding a volume revolved around X or Y axis, the above equation works well except you just have to be careful when you plug in for X or Y.
 
  • #4
33,513
5,195
You could use disks in this problem, but it's more complicated. You will need two integrals to do the problem this way. For y in [0, 1/3], the disks all have a radius of 1. For y in [1/e, 1] the radius of a given disk is the x value on the curve, or +sqrt(-ln(y)).

I should add that in your relevant formula in post 1, pi is missing.
 

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