Calculating volume

  • Thread starter suspenc3
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  • #1
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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?
 

Answers and Replies

  • #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!
 
  • #3
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ohhhhhh..yeah i think i read the question wrong..il try again
 
  • #4
TD
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Ok, a sketch will really help. Also, (tan(x))' = sec²(x), in case you had forgotten :smile:
 
  • #5
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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
 
  • #6
TD
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Why 1-sec²x ?
 
  • #7
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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
 
  • #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.
 
  • #9
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yeah..that makes sense..i just looked at my sketch..I drew it wrong
 
  • #10
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i got something around 3.5...sounds better
 
  • #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:
 
  • #12
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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?
 
  • #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.
 
  • #14
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hrmm..still confused..im never split it into two before
 
  • #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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