Finding the Volume of a solid - Application of Integrals - Exercise that is not clear to understandI've been working on a few exercises and one of them seems not clear, I'm not sure what the author meant in it. Here's the exercise:Find the volume of the solid whose base is the area between the curve y = x 3 and the y axis, from x = 0 to y = 1, considering that his cross sections, taken perpendicular to the y axis, are squares.

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Finding the Volume of a solid - Application of Integrals - Exercise that is not clear to understandI&#039;ve been working on a few exercises and one of them seems not clear, I&#039;m not sure what the author meant in it. Here&#039;s the exercise:Find the volume of the solid whose base is the area between the curve                     y                            =                  x          3                     and the y axis, from   x  =  0 to   y  =  1, considering that his cross sections, taken perpendicular to the y axis, are squares.

polishxcore5z · Accepted Answer

Step 1You will need to review volumes obtained when various cross sectional shapes are used.The typical area-between-curves formula when integrated in x is given as:      ∫          s      t      a      r      t              f      i      n      i      s      h        (      y          t      o      p        −      y          b      o      t      t      o      m        )      d    x to represent vertical cross sectional lines creating the required &quot;area&quot;.When the cross sections are squares leading to a required volume, you will look at expressions of the form      ∫          s      t      a      r      t              f      i      n      i      s      h        (      y          t      o      p        −      y          b      o      t      t      o      m            )    2        d    x, since each cross section is a square with side equal to the distance between the curves.Step 2Since the required volume is with respect to the y-axis, you will need to rewrite the curve in terms of y, i.e.   x  =      y          1              /            3       and look for an integral of the form       ∫          s      t      a      r      t              f      i      n      i      s      h        (      x          r      i      g      h      t        −      x          l      e      f      t            )    2        d    y

Kyle Liu · Accepted Answer

Explanation:From   y  =      x    3   we have   x  =      y    3   and a cross section at position y is a square of side       y    3   and has area   A  =  (      y    3        )    2  , so the volume is:       ∫    0    1    (      y    3        )    2    d  y

Find the volume of the solid whose base is the area between the curve y=x^3 and the y axis, from x=0 to y=1, considering that his cross sections, taken perpendicular to the y axis, are squares.

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