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Improper Integral Volume Around x-axis

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data

    By rotating R=[tex]$\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] about the x axis we obtain a solid with the volume V = ______

    2. Relevant equations



    3. The attempt at a solution
    [tex]$\int _0^{\infty }\frac{dx}{0.6 x+1.7}$[/tex] is divergent
    but what do i do to get the volume? if dont have a and b..
     
  2. jcsd
  3. Mar 11, 2010 #2

    Mark44

    Staff: Mentor

    Break the integral up into two integrals. You can split them anywhere; I'm choosing 1.
    [tex]\int_a^1 \frac{dx}{0.6x + 1.7} + \int_1^b \frac{dx}{0.6x + 1.7}[/tex]

    Both are improper integrals. Take the limit of the first as a --> 0 and the second as b --> infinity.
     
  4. Mar 11, 2010 #3
    Is your integral correct? It appears your integral computes the area of R and not the volume of rotating it about the x-axis. Look up Disc Method in your calculus book for a hint on properly setting up the integral.
     
  5. Mar 11, 2010 #4

    Mark44

    Staff: Mentor

    rs1n, Good point. Slimsta doesn't have the integrand set up correctly to get the volume.
     
  6. Mar 11, 2010 #5
    well the question is give like that..
    [tex]$f(x)=\frac{1}{0.6 x+1.7}$[/tex]
    [tex]$\int _0^t\frac{dx}{0.6 x+1.7}=\left. \frac{\ln (0.6 x+1.7 )}{0.6 }\right| _0^t=$ [/tex] [ln(0.6 t+1.7 )]/0.6 - [ln(1.7 )]/0.6

    and taking the limit as t[tex]$\to \infty$[/tex] we conclude that
    [tex]$\int _0^{\infty }\frac{dx}{0.6 x+1.7}$[/tex] is divergent .

    Therefore the region R[tex]$=\{ (x,y)|x\geq 0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] has infinite area.

    By rotating R[tex]$=\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] about the x axis we obtain a solid with the volume V =___

    so if i use
    [tex]
    \int_a^1 \frac{dx}{0.6x + 1.7} + \int_1^b \frac{dx}{0.6x + 1.7}
    [/tex]
    [tex]
    \int_0^1 \frac{dx}{0.6x + 1.7} + \int_1^\infty \frac{dx}{0.6x + 1.7}
    [/tex]

    ==> [ln(0.6+1.7 )]/0.6 - ln(1.7 )]/0.6] + [infinity - ln(0.6+1.7 )]/0.6]

    how do i get a value for the volume?
     
  7. Mar 11, 2010 #6

    Mark44

    Staff: Mentor

    OK, they're doing two things: 1) determining the area of region R, and 2) determining the volume of the solid of revolution.

    If the graph of f(x) = 1/(.6x + 1.7) is revolved around the x-axis, what integral represents the volume of the solid?

    Hint: This ain't it.
    [tex]\int _0^{\infty }\frac{dx}{0.6 x+1.7}[/tex]

    Draw a sketch of the solid. You're probably going to want to use disks to calculate the volume.

    Note: My previous advice for splitting the integral into two integrals was incorrect. For some reason I was thinking that the integrand was undefined at 0 - it isn't. Your technique of integrating from 0 to t is the right way to go. Sorry for the errant advice.
     
  8. Mar 11, 2010 #7
    i just dont understand what R [tex] $=\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$ [/tex] means.
    my guess would be
    [tex] \int _x^y \frac{dx}{0.6 x+1.7} [/tex] x= 0, y=0.6 ?
    but that doesnt look right
     
  9. Mar 11, 2010 #8

    Mark44

    Staff: Mentor

    That definition just means that R is the set of all points (x, y) for which the x value is >= 0, and the y value is between the x-axis and the curve y = 1/(.6x + 1.7). So R is the region in the first quadrant above the x-axis and below that curve.

    What you need to do is to revolve that region around the x-axis and find the volume of the resulting solid.

    Do you know how to find the volume of a solid of revolution? There are a couple of ways to do it. One way involves disks, and another way involves cylindrical shells. The disk method is probably the way to go here.

    The integral you show DOES NOT give you the volume of the solid. Forget that one; you need a different integral.
     
  10. Mar 11, 2010 #9
    V= 2pi[tex] \int _0^{\infty } yt-yb [/tex]
    V= 2pi[tex] \int _0^{\infty } [1/(0.6 x+1.7)-0 ] dx [/tex]
    isnt that it?
     
  11. Mar 11, 2010 #10

    Mark44

    Staff: Mentor

    No. What's the volume of a disk of radius r and thickness dx?
     
  12. Mar 11, 2010 #11
    oh my bad. pi*r^2
    so pi*[tex] \int _0^{\infty } [(1/(0.6 x+1.7))^2 - 0 ]dx [/tex]
     
  13. Mar 11, 2010 #12

    Mark44

    Staff: Mentor

    Right, and that one you can integrate as you did with the other integral, integrating from 0 to t, say, and then taking the limit as t --> infinity.
     
  14. Mar 11, 2010 #13
    sweet. then i get 3.079!
    solid!
     
  15. Mar 11, 2010 #14

    Mark44

    Staff: Mentor

    No, it's not 0. You're not evaluating your antiderivative correctly.
     
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