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Math Help - Volume Problems

  1. #16
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    Thank you, everything makes sense now.
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  2. #17
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    So for part a I get the volume to be 157.6 Inches cubed, and for part b I still need to figure out the volume of the football using the equation of the parabola, is there a formula I need in order to relate the two equations ?
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  3. #18
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    You can integrate to find volume.
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  4. #19
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    Would my limits be -5.5 and 5.5 ?
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  5. #20
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    You could do 0, 5.5 and multiple by 2. Also, use the volume integral formulas. Regular integration solves area.
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  6. #21
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    volume integral formulas ? where do i find them ? or the right one ?
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  7. #22
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    \displaystyle V=\pi\int_{a}^{b}[f(x)]^2dx
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  8. #23
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    Thank you.
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  9. #24
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    Ok so when I evaluate my Volume I get a really huge number 3000 something which I know isn't right. This is what I evaluate. 2pi* the integral from 0 to 5.5 of ((-21)/2(5.5)^2pi (x^2)+ 21/2pi))^2dx. When I put that in my calculator and integrate I get a huge number. A friend of mine suggested I try a quad regression but I don't know how to do that on the calculator. Please tell me what I did wrong.
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  10. #25
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    I obtained 77 when I integrated it. Show your steps.
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  11. #26
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    I did show you my steps. ok I get for the equation Y = (21/2pi) - (21/2pi(5.5)^2) x^2. and then when I put that into the Volume formula I just move 2pi out front and square f(x) I put the whole function into my calculator and I use my calulator to integrate from 0 to 5.5 I get 3000 something
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  12. #27
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    Never mind good sir. I was plugging it in to the calculator wrong. I get 77 too now. Thank you
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  13. #28
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    \displaystyle 2\pi\int_{0}^{5.5}\left(\frac{21}{2\pi}-\frac{21}{2\pi *5.5^2}*x^2\right)dx=21\left[x]^{5.5}_0-\frac{21}{5.5^2}\left[\frac{x^3}{3}\right] _0^{5.5}

    \displaystyle = 21*5.5-\frac{21}{5.5^2}\left[\frac{5.5^3}{3}\right]=77
    Last edited by dwsmith; November 29th 2010 at 04:14 PM.
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  14. #29
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    Yes yes thank you once again.
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