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Math Help - Calculus area problem

  1. #1
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    Calculus area problem

    Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

    The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are
    a) circular disks with diameters running from the curve y = tanx to the curve y = sec x
    b) squares whose bases run from the curve y = tan x to the curve y = sec x
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by turtle View Post
    Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

    The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are
    a) circular disks with diameters running from the curve y = tanx to the curve y = sec x
    As sec(x)>tan(x) for x in (-pi/3, pi/3), the volume of the solid is:

    int pi (sec(x)-tan(x))^2 /4 dx = sqrt(3) pi - pi^2/6 ~=3.7965

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by turtle View Post
    Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

    The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are

    b) squares whose bases run from the curve y = tan x to the curve y = sec x
    Same as part (a) except instead of integrating pi (sec(x)-tan(x))^2/4 from
    -pi/3 to pi/3 you will integrate (sec(x)-tan(x))^2/4 from -pi/3 to pi/3.

    RonL
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  4. #4
    Newbie pocketasian's Avatar
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    This is an old question, but how would you find this analytically for part a?

    For the volume, so far I have:
    (pi/4) times the integral from -pi/3 to pi/3 of (2(secx)^2 - 2secxtanx - 1)dx

    I tried using u-substitution, making u=secx and du=secxtanxdx, but I got stuck. Will someone help me solve this by hand? Thanks.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by pocketasian View Post
    This is an old question, but how would you find this analytically for part a?

    For the volume, so far I have:
    (pi/4) times the integral from -pi/3 to pi/3 of (2(secx)^2 - 2secxtanx - 1)dx

    I tried using u-substitution, making u=secx and du=secxtanxdx, but I got stuck. Will someone help me solve this by hand? Thanks.
    [\sec(x)]^2 is a standard integral, it integrates to \tan(x)+c. For \sec(x)\tan(x) replace the trig functions by \sin 's and \cos 's and you are left with the derivative of 1/\cos(x).

    CB
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