Computing volume using shell resulted in negative value?The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...Use cylindrical shells to compute the volume of y = x 2 and y = 2 − x 2 , revolved about x = 2. x 2 = 2 − x 2 → 2 x 2 = 2 → 2 x 2 2 = 2 2 → x 2 = 1 x = 1 , x = − 1Radius is r = 2 − xHeight is x 2 − ( 2 − x 2 )Finding the Integral, ∫ ¬ 1 1 2 π ( 2 − x ) ( x 2 − ( 2 − x 2 ) ) d x = 2 π ( 4 x 3 3 − 4 x − 2 x 4 3 + 2 x 2 − x 2 + x 4 6 ) + CFinding the limits, lim x → − 1 + ⁡ 2 π ( 4 x 3 3 − 4 x − 2 x 4 3 + 2 x 2 − x 2 + x 4 6 ) = 2 π ( 4 ( − 1 ) 3 3 − 4 ( − 1 ) − 2 ( − 1 ) 4 3 + 2 ( − 1 ) 2 − ( − 1 ) 2 + ( − 1 ) 4 6 ) = 19.897 lim x → 1 − ⁡ 2 π ( 4 x 3 3 − 4 x − 2 x 4 3 + 2 x 2 − x 2 + x 4 6 ) = 2 π ( 4 ( 1 ) 3 3 − 4 ( 1 ) − 2 ( 1 ) 4 3 + 2 ( 1 ) 2 − ( 1 ) 2 + ( 1 ) 4 6 ) = − 13.614And finally, computing the volume from the obtained limits V = − 13.614 − 19.897 = − 33.510.So, if someone could let me know if (or where) I went wrong, that would be great.

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Computing volume using shell resulted in negative value?The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...Use cylindrical shells to compute the volume of   y  =      x    2   and   y  =  2  −      x    2  , revolved about   x  =  2.      x    2    =  2  −      x    2    →  2      x    2    =  2  →            2              x        2              2    =      2    2    →      x    2    =  1  x  =  1  ,  x  =  −  1Radius is   r  =  2  −  xHeight is       x    2    −  (  2  −      x    2    )Finding the Integral,       ∫          ¬      1        1    2      π    (  2  −  x  )  (      x    2    −  (  2  −      x    2    )  )  d  x  =  2      π    (            4              x        3              3    −  4  x  −            2              x        4              3    +  2      x    2    −      x    2    +            x      4        6    )  +  CFinding the limits,       lim          x      →      −      1      +            ⁡    2      π    (            4              x        3              3    −  4  x  −            2              x        4              3    +  2      x    2    −      x    2    +            x      4        6    )  =  2      π    (            4      (      −      1              )        3              3    −  4  (  −  1  )  −            2      (      −      1              )        4              3    +  2  (  −  1      )    2    −  (  −  1      )    2    +            (      −      1              )        4              6    )  =  19.897      lim          x      →      1      −            ⁡    2      π    (            4              x        3              3    −  4  x  −            2              x        4              3    +  2      x    2    −      x    2    +            x      4        6    )  =  2      π    (            4      (      1              )        3              3    −  4  (  1  )  −            2      (      1              )        4              3    +  2  (  1      )    2    −  (  1      )    2    +            (      1              )        4              6    )  =  −  13.614And finally, computing the volume from the obtained limits   V  =  −  13.614  −  19.897  =  −  33.510.So, if someone could let me know if (or where) I went wrong, that would be great.

slapadabassyc · Accepted Answer

Explanation:In the region   −  1  ≤  x  ≤  1,   2  −      x    2    &amp;gt;      x    2   (the   2  −      x    2   curve is above the       x    2   curve). So the height is   (  2  −      x    2    )  −      x    2  , not       x    2    −  (  2  −      x    2    ).

Kade Reese · Accepted Answer

Step 1The expression of the volume should be      ∫          −      1        1    2      π    (  2  −  x  )  (  (  2  −      x    2    )  −      x    2    )  d  xStep 2Because,   2  −      x    2    ≥      x    2      −  1  ≤  x  ≤  1

Use cylindrical shells to compute the volume of y=x^2 and y=2-x^2, revolved about x=2.

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