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How to find the radius in each of these integrals?

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Given: y = sqrt(x), y=0, x=3

    Find the volume of the solid bounded by these functions, revolved around B) y-axis and C) line x=3. (Disk method)

    2. Relevant equations
    Disk method of finding volume using π ∫ r2dy

    3. The attempt at a solution

    Ok so, the part of the problem that I am having trouble on is how to find the radius. I know that both integrals will be with respect to dy with the limits being 0 and sqrt(3) and I understand that these two integrals SHOULD be different, but I'm not sure what that difference is in terms of the radius in πr2.

    Please refer to the solution to B and C given in this link (http://www.calcchat.com/book/Calculus-10e/7/2/11/). My question is, why is the radius [32 - (y2)2] in B and why is the radius [3-y2]2 in C. My problem is understanding why the power of 2 is distributed differently and how I'm supposed to interpret that based on the problem.

    Thanks,
     
  2. jcsd
  3. Oct 6, 2015 #2

    BvU

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    Hello Celestor, welcome to PF :smile: !

    Your question is a bit strange. The radius itself isn't [ 32-(y2)2 ] in b. What do you get when you rotate the shaded area around the y-axis ?

    [edit] I mean the dark shaded rectangular area....​

    In (c) you rotate around the line x = 3, so what do you get when you rotate the shaded area around the dashed line ?


    [edit] cute that you get to see the exercise worked out at the website. But doesn't that prevent you from first trying to find your own path towards a solution ?​
     
    Last edited: Oct 6, 2015
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