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Finding the Volume of a Solid

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y =4.
    $$y=\frac 3 {1+x},~ y=0,~ x=0,~x=3$$

    2. Relevant equations

    $$V= \int_a^b ([R(x)]^2-[r(x)]^2)dx$$

    3. The attempt at a solution

    I understand how to use the equation, but I don't know how to find all the components needed to plug into the equation. R(x) and r(x) are both the inner and outer radius, however, I don't understand why R(x)=4 and ##r(x)=4-\frac 3 {1+x}##. I understand that (a,b) are (0,3) since that is what x equals. I'm just having trouble understanding the radius.

    This website showed how to solve the problem, but it doesn't really explain each step.
    http://www.calcchat.com/book/Calculus-ETF-6e/7/2/17/
     
  2. jcsd
  3. Feb 20, 2017 #2

    Student100

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    Did you draw a picture? That's most helpful for these problems. Try to actually draw the solid after rotation.

    So the solid you're trying to find the volume of is rotated about the line y =4. Now, what you're trying to determine is an inner and outer radius from the axis of rotation. So the outer radius is the bottom function, in this case it's just the x axis (the line y=0). The distance of which is a constant 4 from the axis of rotation. The inner radius varies with the function provided. So it changes as you move along the path of integration. Try to set up a picture now, draw the resultant solid, then try to see if you can visualize the radii.
     
  4. Feb 20, 2017 #3
    Here is my drawing, I'm not sure I drew it correctly. The inner radius is the hole, correct? So wouldn't the inner radius be what is in between the bottom drawing and the top? problem 17.jpg
     
  5. Feb 20, 2017 #4

    Student100

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    So for the artistically impaired (like me), I would do this..

    graph.png



    Do you see how it's mirrored about the axis of rotation?
     
  6. Feb 20, 2017 #5
    I see where the outer radius, however, I don't understand where the inner radius would be.
    #17.jpg
     
  7. Feb 20, 2017 #6

    Student100

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    Draw a few lines from the axis of rotation to curve ##\frac{3}{1+x}##. You can also then see why it needs to be ##4- \frac{3}{1+x}## then.

    Look at the case x=0 and x=3 specifically.
     
  8. Feb 20, 2017 #7
    Is it because the inner radius is below y=4 and what is below that is ##\frac{3}{1+x}##. So it would be ##4- \frac{3}{1+x}##? FullSizeRender.jpg
     
  9. Feb 20, 2017 #8

    Student100

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    The inner radius is the curve that forms the part of the solid closest to the axis of rotation, while the outer radius is the one furthest away. It's not easy for me to put into words..

    Try reading: http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx see if that helps with their better pictures. Pay close attention to example 3.
     
  10. Feb 20, 2017 #9

    LCKurtz

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    You have shaded an incorrect area that is being rotated. Shade the region under ##y=\frac 3 {1+x}## above the ##x## axis between ##x=0## and ##x=3##. If you are thinking about the wrong area no wonder you don't get the radius correct.
     
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