Finding the volume by Shell method: y = x 2 , y = 2 − x 2 , y = x 2 , y = 2 − x 2 about the line x = 1.what I get from this after graphing is: 2 π ∫ ( 1 − x ) ( 2 − 2 x 2 ) d xwhich becomes: 2 π ∫ ( 2 − 2 x 2 − 2 x + 2 x 3 ) d xintegrating that I get: 2 π [ 2 x − 2 3 x 3 − x 2 + 1 2 x 4 ] from 0 to 1My answer is 5 3 π but my book says the answer is: 16 3 πCould someone tell me where I went wrong? Was it the upper lower bounds?

Question

Finding the volume by Shell method:   y  =      x    2    ,  y  =  2  −      x    2  ,   y  =      x    2    ,  y  =  2  −      x    2   about the line   x  =  1.what I get from this after graphing is:  2  π  ∫  (  1  −  x  )  (  2  −  2      x    2    )  d  xwhich becomes:   2  π  ∫  (  2  −  2      x    2    −  2  x  +  2      x    3    )  d  xintegrating that I get:  2  π  [  2  x  −      2    3        x    3    −      x    2    +      1    2        x    4    ] from 0 to 1My answer is       5    3    π  but my book says the answer is:       16    3    πCould someone tell me where I went wrong? Was it the upper lower bounds?

Monastro3n · Accepted Answer

Explanation:The limits of the integration is given by where the lines cross each other:      x    2    =  2  −      x    2    ⟺  x  =  ±  1so the limits shold be from -1 to 1.

Haley Madden · Accepted Answer

Explanation:The region in the xy-plane bounded by the graphs   y  =      x    2   and   y  =  2  −      x    2   is symmetric about the y-axis, so the projection of this region onto the x-axis should be a symmetric interval of the form [-a,a]. In this case the points of intersection of the graphs have x-coordinates   x  =  ±  1. Thus, your limits of integration should really be from   x  =  −  1 to   x  =  1, not 0 to 1.