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Math Help - 2 Volume Problems

  1. #1
    Junior Member
    Joined
    Oct 2006
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    2 Volume Problems

    If anyone could help with the following problems, I would greatly appreciate it. I will explain what I have done beneath the problems in case I am going in the right direction.

    Find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis.
    1. y = sec x , y = square root of 2 , -Π/4 ≤ x ≤ Π/4
    I set R = square root of 2 and r = sec x tan x
    I wasn't able to integrate sec^2 x tan^2 x (if I was on the
    right track)
    2. y = x^2 + 1 , y = x + 3
    I set R = (x+3) - (x^2 +1) and r = 1. I don't know how to find the
    intervals of integration.
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  2. #2
    MHF Contributor
    Joined
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    Quote Originally Posted by turtle View Post
    If anyone could help with the following problems, I would greatly appreciate it. I will explain what I have done beneath the problems in case I am going in the right direction.

    Find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis.
    1. y = sec x , y = square root of 2 , -Π/4 ≤ x ≤ Π/4
    I set R = square root of 2 and r = sec x tan x
    I wasn't able to integrate sec^2 x tan^2 x (if I was on the
    right track)
    2. y = x^2 + 1 , y = x + 3
    I set R = (x+3) - (x^2 +1) and r = 1. I don't know how to find the
    intervals of integration.
    What do you mean by R and r?
    I assume you are using vertical washers for the integrations
    Thus, dV = pi(R^2 -r^2)dX.

    If that is so, then why is the r for the Problem #1 be secXtanX?
    That r should be secX.
    So, the dV is
    dV = pi[(sqrt(2))^2 -sec^2(X)]dx
    dV = pi[2 -sec^2(X)]dx

    Find the boundaries of dx. They are the x-values at the intersections of the horizontal line and the secant line:
    secX = sqrt(2)
    X = arcsec(sqrt(2)) = pi/4 or -pi/4
    Therefore, the boundaries of dx are from -pi/4 to pi/4.
    So,
    V = (pi)INT.(-pi/4 -->pi/4)[2 -sec^2(X)]dX
    V = pi[2X -tanX]|(-pi/4 --> pi/4)
    V = pi[2(pi/4) -tan(pi/4) -2(-pi/4) +tan(-pi/4)]
    V = pi[pi/2 -1 +pi/2 +(-1)]
    V = pi[pi -2]
    V = 3.5864 cubic units --------------answer.

    =========================================
    For the #2.

    R = x+3
    r = x^2 +1

    dV = pi(R^2 -r^2)dx
    dV = pi[(x+3)^2 -(x^2 +1)^2]dx
    dV = pi[x^2 +6x +9 -(x^4 +2x^2 +1)]dx
    dV = pi[x^2 +6x +9 -x^4 -2x^2 -1]dx
    dV = pi[8 +6x -x^2 -x^4]dx --------------**

    Boundaries for dx,
    x+3 = x^2 +1
    0 = x^2 -x +1 -3
    x^2 -x -2 = 0
    (x-2)(x+1) = 0
    x = 2 or -1
    So boundaries of dx are from x = -1 to x=2

    Hence,
    V = (pi)INT.(-1 --> 2)[8 +6x -x^2 -x^4]dx
    V = pi[8x +3x^2 -(1/3)x^3 -(1/5)x^5]|(-1 --> 2)
    V = pi[16 +3(4) -(1/3)(8) -(1/5)(32) -[-8 +3 -(1/3)(-1) -(1/5)(-1)]]
    V = pi[28 -8/3 -32/5 +8 -3 -1/3 -1/5]
    V = 23.4pi
    V = 75.513 cu. units -------------------answer.
    Last edited by ticbol; February 11th 2007 at 03:13 AM. Reason: Wrong integration at #2
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