Finding volume by rotating cos x and sin x about y = − 1, x ∈ [ 0 , π 4 ) I'm working on a problem that is asking for the volume of y = cos ⁡ x when rotated about the line y = 1, with a restricted domain of [ 0 , π 2 ] . The range is [0, 2].There's a problem I'm working on that is asking for the volume of y = cos ⁡ x and y = sin ⁡ x when rotated about line y = − 1 with a domain of [ 0 , π 4 ) I don't even know where to begin. First off; which method would be most effective in this case? A shell, washer or disc?

Question

Finding volume by rotating cos x and sin x about   y  =  −  1,   x  ∈            [        0  ,      π    4              )      I&#039;m working on a problem that is asking for the volume of   y  =  cos  ⁡  x when rotated about the line   y  =  1, with a restricted domain of             [        0  ,      π    2              ]      . The range is [0, 2].There&#039;s a problem I&#039;m working on that is asking for the volume of   y  =  cos  ⁡  x and   y  =  sin  ⁡  x when rotated about line   y  =  −  1 with a domain of             [        0  ,      π    4              )      I don&#039;t even know where to begin. First off; which method would be most effective in this case? A shell, washer or disc?

honotMornne · Accepted Answer

Step 1I assume that for the first problem, you are rotating the region below   y  =  cos  ⁡  x, above the x-axis, from   x  =  0 to   x  =      π    2  .Both Slicing and the Method of Shells work. Perhaps Slicing is a little easier.Draw a picture. Make a slice perpendicular to th x-axis &quot;at&quot; x.The cross-section is a &quot;washer,&quot; that is, a circle with a circular hole in it. The radius of the outer circle is 1, and the radius of the inner circle is   1  −  y, that is,   1  −  cos  ⁡  x. Thus the area A(x) of cross-section is given by  A  (  x  )  =  π      (          1      2        −    (    1    −    cos    ⁡    x          )      2        )    .Step 2It follows that the volume of the solid is      ∫    0          π              /            2        π      (          1      2        −    (    1    −    cos    ⁡    x          )      2        )      d  x  .Before integrating, it will be useful to simplify the integrand. You will need, among other things, to integrate       cos    2    ⁡  x. There are several ways to do it. One standard one is to use the fact that       cos    2    ⁡  x  =            cos      ⁡      (      2      x      )      +      1        2  .The second problem has a very similar structure.

Finding volume by rotating cosx and sinx about y=-1, x in [0, pi/4)

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