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Calc II

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid which is obtained by revolving the discs [tex](x - 1)^2 + y^2 \le 1[/tex]

    3. The attempt at a solution

    I say using cylinders would be easiest so,
    [tex]\int_a^b{2\pi r(x)h(x) dx} = \int_0^2{2\pi x\sqrt{1-(x-1)^2} dx}[/tex]

    Any hints on how to get started with this integral?

    ---

    1. The problem statement, all variables and given/known data

    Find the arclength of the curve [itex]y = e^x[/itex] with [itex]0 \le x \le ln 2[/itex]

    3. The attempt at a solution
    [tex]y' =e^x[/tex]
    so I need to solve
    [tex]\int_a^b{\sqrt{1+(y')^2}dx} = \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}[/tex]

    Any hints?
     
    Last edited: Jan 16, 2007
  2. jcsd
  3. Jan 15, 2007 #2
    [tex]\int{\sqrt{1+e^{2x}}[/tex]

    this can be written as [tex]{1+e^{2x}}^(1/2)[/tex]

    then to integrate you add one to the power...etc

    does this help?
     
  4. Jan 15, 2007 #3
    wait that came up wrong i meant:

    this can be written as (1+e^(2x))^(1/2)
     
  5. Jan 16, 2007 #4
    I can't do what you said as it's a composite function...

    Any other ideas?
     
  6. Jan 17, 2007 #5

    Gib Z

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    sara_87, when doing exponents longer than one character, uses { } instead of ( ). It seems you just forgot, since you did it the first time with the 2x :D
     
  7. Jan 17, 2007 #6

    Gib Z

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    o, cscott, would It help If You had it in this form? (Hopefully you know integration by parts ).

    Make a u substitution
    [tex]u=e^{2x}[/tex]
    [tex]\frac{du}{dx}=2e^{2x}[/tex]
    Solve for dx
    [tex]dx=\frac {du}{2e^{2x}}=\frac{du}{2u}[/tex]
    Now you just need to do the rest.

    [tex]\int (u+1)^{\frac{1}{2}} 2u du[/tex]

    Have fun :D
     
    Last edited: Jan 17, 2007
  8. Jan 17, 2007 #7
    question 1)Not that I want to ruin your fun here, but by revolving discs do you mean a sphere? If I draw a graph of a revolving disc unless it's simply rotating about it's axis which gives a circle, wouldn't I end up with a sphere? in other words [tex]\frac{4}{3}\pi r^3[/tex]
     
    Last edited: Jan 17, 2007
  9. Jan 17, 2007 #8

    cristo

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    You made a slight mistake; the integral is [tex]\int \frac{(u+1)^{\frac{1}{2}}}{2u} du[/tex]
     
  10. Jan 17, 2007 #9
    Sorry, it should say "revolve the disc defined by ___ around the y-axis" which I think calls for cylinders to make it easiest and wouldn't be a sphere.
     
    Last edited: Jan 17, 2007
  11. Jan 17, 2007 #10
    Oh you mean revovle a cyclinder around the y axis. I suggest you copy out the question again exactly as it is on paper, as your original question is a little misleading. I thought you meant spinning a circle through 360 degrees.
     
  12. Jan 17, 2007 #11

    Gib Z

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    cristo- ty for spotting this, my brain seems to be farting alot recently...

    cscott- either way, I still think integration by parts helps
     
  13. Jan 18, 2007 #12
    The first question I'm assuming your talking about a toroidal. which is found from the curve y=e^x. You may want to look up toroidal volumes.

    Hint: you should end up with the second question integral, part 1 if you have done it correctly.

    i.e. [tex]\int_a^b{\sqrt{1+(y')^2}dx= \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}[/tex]

    The second I can only say was too complicated for the level I'm at although once I saw how it was done I could follow the steps easilly enough(I asked Dr.E.P.A Bailey. A colleague to solve it basically) Otherwise known as cheating :smile:

    let [tex]u^2=1+e^{2x}[/tex] and then substitute.

    then put in the form of partial fractions.

    [tex] \frac {f(u)}{g(u)}[/tex]

    where f(u) and g(u) are polynomials.

    You should end up with this:-

    [tex]\int_0^{ln 2} 1+\frac {1}{2(u-1)}. \frac {-1}{2(u+1)}[/tex]

    from here it should be fairly straightforward if you express the answer in the form of natural logs.

    Hope that helps.
     
    Last edited: Jan 18, 2007
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