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Math Help - help with volume: disks and washers

  1. #1
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    help with volume: disks and washers

    Say I am given the following boundaries:

    y=e^10x, y=1 , y=3.

    I know how to find the volume of things like

    revolved about y=1,y=e^30,y=-4


    But what if I am told to find the volume if:

    revolved bout the y-axis or the x-axis


    how would I find these volumes?
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  2. #2
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    Quote Originally Posted by gammaman View Post
    Say I am given the following boundaries:

    y=e^10x, y=1 , y=3.

    I know how to find the volume of things like

    revolved about y=1,y=e^30,y=-4
    Well, I don't! Because you are missing a left boundary. Did you intend x= 0 as a boundary?



    But what if I am told to find the volume if:

    revolved bout the y-axis or the x-axis


    how would I find these volumes?
    The first of those is easy because now you can assume that x= 0 is a boundary. Imagine a horizontal line, representing a thin disk as it is rotated around the y-axis. Its boundary will be a circle having radius from x= 0 to x= ln(y)/10 (solving y= e^{10x} for x). Such a circle has area \pi (ln(y))^2/100 and, taking the disk to have thickness dy, the disk will have volume [tex]\frac{\pi}{100}(ln(y))^2dy[tex] and the whole figure will have volume [tex]\frac{\pi}{100}\int_1^3 (ln(y))^2dy[/itex]

    For rotating around the x-axis, again, you will need to give some left boundary, perhaps the y-axis, x= 0. Now, the simplest thing to do is to look at separate parts: The upper boundary, y= 3, is a horizontal line and rotating around the x-axis gives a cylinder with radius 3 and height ln(10)/3 (where y= e^{10x} crosses y= 3. That cylinder has volume 3ln(10)\pi. The lower boundary is y= e^{10x} and if we rotate the area below that around the x-axis, each thin disk will have radius e^{10x} so area \pi e^{20x} and volume \pi e^{20x}dx so the total volume is \pi\int_0^{\frac{ln(3)}{10}}dx. That is NOT the volume you want- the volume you want is the volume of the cylinder minus that integral. (This is exactly the same thing as using the washer method, doing that subtraction before the integral.)
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