Finding value to calculate volumeGiven D = [ ( x ; y ) ∈ R 2 : 1 ≤ y ≤ a x 2 + 1 , 0 ≤ x ≤ 2 / a ] , 0 &lt; a let W be the region obtained by rotating D around Y axis.A) Find the volume of WB) Find, if possible, the a values ϵ ( 0 ; + ∞ ) so that the volume of W is a minimum, and a maximumC) Find, if possible, the a values ϵ ( 1 / 3 ; 3 ) so that the volume of W is a minimum, and a maximumWell, what I've done so far is finding the inner and outer radius to calculate the volume in terms of Y. The inner radius would be be y − 1 a and the outer would be 4 + a a , which is a evaluated in the parabola. Then, the volume would be ∫ 1 4 + a a ( y − 1 a ) 2 − ( 4 + a a ) 2 d x.And that's the function that I have to differentiate to find its maxima and minima, which, after differentiating, is 1/a. Is ok what am I doing? How can I go on?

Question

Finding value to calculate volumeGiven   D  =  [      (    x    ;    y    )    ∈                  R            2        :    1    ≤    y    ≤    a          x      2        +    1    ,    0    ≤    x    ≤    2          /        a    ]  ,  0  &amp;lt;  a let W be the region obtained by rotating D around Y axis.A) Find the volume of WB) Find, if possible, the a values   ϵ  (  0  ;  +  ∞  ) so that the volume of W is a minimum, and a maximumC) Find, if possible, the a values   ϵ  (  1      /    3  ;  3  ) so that the volume of W is a minimum, and a maximumWell, what I&#039;ve done so far is finding the inner and outer radius to calculate the volume in terms of Y. The inner radius would be be                     y        −        1            a       and the outer would be             4      +      a        a  , which is a evaluated in the parabola. Then, the volume would be       ∫    1                  4        +        a            a        (                    y        −        1            a            )    2    −  (            4      +      a        a        )    2      d  x.And that&#039;s the function that I have to differentiate to find its maxima and minima, which, after differentiating, is 1/a. Is ok what am I doing? How can I go on?

Malcolm Mcbride · Accepted Answer

Explanation:The smallest value of y is 1 and its largest value is   f      (          2      a        )    =            4      +      a        a  . For each y in that interval, the possible values of x go from                     y        −        1            a       to       2    a  . Therefore, that volume is equal to   π      ∫    1                            4          +          a                a                        (              2        a            )        2    −            (                                    y            −            1                    a                    )        2          d    y  =  π      (          8              a        3              +          4              a        2              +          1              2        a              )    .

Given D=[(x;y) in mathbb{R}^{2}:1 leq y leq ax^{2}+1,0 leq x leq 2/a],0, let <a W be the region obtained by rotating D around Y axis.

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