Trouble finding the volume using the shell methodA region is bounded by the line y = 3 x + 4 and the parabola y = x 2 and is rotated about the line x = 4.First I found the limits of integration by finding the points of intersection. They are (-1,1) and (4,16). I then found the radius from the axis of rotation to be r = 4 − x and the height to be h = 3 x + 4 − x 2 .I used the Volume formula ∫ a b 2 π ( r a d i u s ) ( h e i g h t ) d x. V = ∫ − 1 4 2 π ( 4 − x ) ( 3 x + 4 − x 2 ) d x 2 π ∫ − 1 4 ( x 3 − 7 x 2 + 8 x + 16 ) d xEvaluating this integral I get the volume to be 280 π − 1055 6 but this is no where close to the actual volume. I believe that both my radius and height are right and that the formula is right. I've redid the problem a couple of times and ended up with the same answer, what am I missing here?

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Trouble finding the volume using the shell methodA region is bounded by the line   y  =  3  x  +  4 and the parabola   y  =      x    2   and is rotated about the line   x  =  4.First I found the limits of integration by finding the points of intersection. They are (-1,1) and (4,16). I then found the radius from the axis of rotation to be   r  =  4  −  x and the height to be   h  =  3  x  +  4  −      x    2  .I used the Volume formula       ∫          a              b        2  π  (  r  a  d  i  u  s  )  (  h  e  i  g  h  t  )  d  x.  V  =      ∫          −      1              4        2  π  (  4  −  x  )  (  3  x  +  4  −      x    2    )  d  x  2  π      ∫          −      1              4        (      x    3    −  7      x    2    +  8  x  +  16  )  d  xEvaluating this integral I get the volume to be   280  π  −            1055      6       but this is no where close to the actual volume. I believe that both my radius and height are right and that the formula is right. I&#039;ve redid the problem a couple of times and ended up with the same answer, what am I missing here?

Raelynn Johnson · Accepted Answer

Step 1Everything is fine in what you have; the only thing missing is the actual evaluation of the definite integral. I get       2    π                  [                              x            4                    4                −                              7                          x              3                                3                +                              8                          x              2                                2                +        16        x        ]                    −        1                    4                  2    π          [              (        64        −                              7            (            64            )                    3                +        64        +        64        )            −              (                  1          4                +                  7          3                +                  8          2                −        16        )            ]      Step 2From there, it&#039;s just a matter of simple algebra. I get                     625        π            6

A region is bounded by the line y=3x+4 and the parabola y=x^{2} and is rotated about the line x=4.

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