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Homework Help: Calculating volume

  1. Jun 1, 2006 #1
    Hi, im having a little trouble with this question, im trying to find the volume of the solid after revolving it.

    y=sec x, y=1, x=-1, x=1, about the x-axis

    what I did:

    [tex]A(x)= \frac{ \pi}{(cosx)^2}[/tex]

    and then I integrated from -1 to 1...(my integral came out to [tex]\frac{-1}{sinx}[/tex])
    and I got an answer that came out negative..

    can anyone help?
     
  2. jcsd
  3. Jun 1, 2006 #2

    TD

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    Have you made a sketch of the problem?

    Don't forget that you're only considering the area between y = sec(x) and y = 1!
     
  4. Jun 1, 2006 #3
    ohhhhhh..yeah i think i read the question wrong..il try again
     
  5. Jun 1, 2006 #4

    TD

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    Ok, a sketch will really help. Also, (tan(x))' = sec²(x), in case you had forgotten :smile:
     
  6. Jun 1, 2006 #5
    ok..i tried it again..but im still getting it wrong..

    heres what i have:

    [tex]A(x)=\pi(1-(secx)^2)[/tex]

    and then i integrated from -1 to 1 and then i ended up with somethign around 14
     
  7. Jun 1, 2006 #6

    TD

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    Why 1-sec²x ?
     
  8. Jun 1, 2006 #7
    im not really sure how to do this one..dont you do this..A=pi(outer radius)^2 - pi(inner radius)^2..i think its called the washer method
     
  9. Jun 1, 2006 #8

    TD

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    Yes, but on [-1,1], which one is the 'outer radius'? I believe sec(x) >= 1 there...

    Remember that sec(x) = 1/cos(x) and cos(x) is bounded between [-1,1]. So without looking at the sign, in absolute value, |sec(x)| will never be smaller than 1.
     
  10. Jun 1, 2006 #9
    yeah..that makes sense..i just looked at my sketch..I drew it wrong
     
  11. Jun 1, 2006 #10
    i got something around 3.5...sounds better
     
  12. Jun 1, 2006 #11

    TD

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    I'm getting the same :smile:

    [tex]
    \pi \int\limits_{ - 1}^1 {\sec ^2 x - 1dx} = 2\pi \left( {\tan 1 - 1} \right) \approx 3.5
    [/tex]

    A good sketch will lead you to the answer :wink:
     
  13. Jun 1, 2006 #12
    ok, thanks, IM having trouble with another one..Im having rouble seting it up. y=x and y=0, x=2, x=4 about x=1

    I drew the two lines..its a pretty simple graph but what does it mean to revolve it around x=1?and do i still have to do the outer radius-inner radius?
     
  14. Jun 1, 2006 #13

    TD

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    The formula for the volume of a solid of revolution, rotated about the x-axis was:

    [tex]
    \pi \int\limits_a^b {y^2 dx}
    [/tex]

    Here, y = f(x). By using inner/outer radii and substracting, you could use this formula as well when the rotation axis was parallel to the x-axis.
    Now, for revolving about the y-axis, or a line parallel to that such as x = 1, the formula logically becomes:

    [tex]
    \pi \int\limits_a^b {x^2 dy}
    [/tex]

    And here, x = f(y).

    In your problem, I think you'll have to split it in two integrals.
     
  15. Jun 1, 2006 #14
    hrmm..still confused..im never split it into two before
     
  16. Jun 1, 2006 #15

    TD

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    Well, your 'outer radius' will be x = 4 all the time. But the inner radius will change. Between y = 0 and y = 2, the inner radius is x = 2. But from y = 2 till y = 4, the inner radius will be y = x. If you have a good sketch, you should be able to see this :smile:
     
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