Volume of revolved solid using shell method: finding heightThe problem that I am working with is:Find the volume of the solid of revolution formed by rotating the region R bounded by y = 4 + x 2 , x = 0 , y = 0 , a n d x = 1 about the line y = 10.I have the following so far (using the shell method): V = ∫ a b 2 π r h d y r = 10 − y c = 2 π ( 10 − y ) h = ? V = 2 π ∫ 0 1 ( 10 − y ) h d yI've been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.Would the height just be y − 4 or would it be 10 − y − 4 and why?

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Volume of revolved solid using shell method: finding heightThe problem that I am working with is:Find the volume of the solid of revolution formed by rotating the region R bounded by   y  =  4  +      x    2    ,    x  =  0  ,    y  =  0  ,    a  n  d    x  =  1  about the line   y  =  10.I have the following so far (using the shell method):  V  =      ∫    a    b    2  π    r    h    d  y    r  =  10  −  y    c  =  2  π  (  10  −  y  )    h  =  ?  V  =  2  π      ∫    0    1    (  10  −  y  )  h  d  yI&#039;ve been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.Would the height just be       y    −    4   or would it be   10  −      y    −    4   and why?

pereishen9g · Accepted Answer

Step 1We will assume that you really want to use the Shell Method. In that case, we will look at a thin strip of width &quot;dy&quot; at height y, and rotate it about   y  =  10.The complication is that the thin strip has a different shape from   y  =  0 to   y  =  4 than it does from   y  =  4 to   y  =  5.From   y  =  0 to   y  =  4, the radius is   10  −  y, and the &quot;height&quot; of the cylindrical shell is 1. Thus the volume obtained by rotating that part is       ∫          y      =      0        4    2  π  (  10  −  y  )  (  1  )    d  y  .Step 2From   y  =  4 to   y  =  5, the radius is still   10  −  y, but the &quot;height&quot; of the cylindrical shell is   1  −  x, that is,   1  −      y    −    4  . The required integral is      ∫          y      =      4        5    2  π  (  10  −  y  )  (  1  −      y    −    4    )    d  y  .Calculate the two integrals, and add.Remark: For this problem, the Method of Washers is less work. Take a cross-section &quot;at&quot; x perpendicular to the x-axis. The outer radius is 10, and the inner radius is       ∫          y      =      4        5    2  π  (  10  −  y  )  (  1  −      y    −    4    )    d  y  .

Find the volume of the solid of revolution formed by rotating the region R bounded by y=4+x^2, x=0, y=0, and x=1 about the line y=10.

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