PLEASE HELP

[rafeeki92]

the base of a solid is the region enclosed by the curve x = 2(square root of y) and the lines x + y = 0 and y = 9. find the volume of the solid if all the plane sections perpendicular to the y axis are squares having a diagonal with one endpoint on the line x + y = 0 and the other endpoint on the curve x = 2 (square root of y)


I have no idea what to do...

 

[soroban]

Hello, rafeeki92!

\(\displaystyle \text{The base of a solid is the region enclosed by the curve }\,x \:=\: 2\sqrt{y}\,\text{ and the lines }\,x + y \:=\: 0\,\text{ and }\,y \:=\: 9.\)

\(\displaystyle \text{Find the volume of the solid if all the plane sections perpendicular to the }y\text{-axis}\)
\(\displaystyle \text{are squares having a diagonal with one endpoint on the line }x + y \:=\: 0\,\text{ and the other endpoint on the curve }x \:=\:2\sqrt{y}\)

First, sketch the region . . .

\(\displaystyle \text{The graph of }\,x \:=\:2\sqrt{y}\,\text{is the right half of the parabola: }\,y \:=\:\tfrac{1}{4}x^2\)

\(\displaystyle \text{The graph of }\,x+y\:=\:0\,\text{ is the line through the origin with slope -1.}\)

   (-9,9)         |         (6,9)
      o - - - - - + - - - - - o
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                  |

\(\displaystyle \text{The right function is: }\:x_{_R} \:=\:2\sqrt{y}\)

\(\displaystyle \text{The left function is: }\:x_{_L} \:=\:-y\)


\(\displaystyle \text{The diagonal of a square cross-section is between the two functions.}\)

              *
            *   *   s
          *       *
        *     d     *
      o - - - - - - -o
        *           *
          *       *
            *   *
              *

\(\displaystyle \text{The length of the diagonal is: }\:d \;=\;x_{_R}}- x_{_L}}\)

\(\displaystyle \text{The side of the square is: }\:s \;=\;\frac{d}{\sqrt{2}} \;=\;\frac{x_{_R} - x_{_L}}{\sqrt{2}}\)

\(\displaystyle \text{The area of the square is: }\;s^2\;=\;\left(\frac{x_{_R} - x_{_L}}{\sqrt{2}}\right)^2 \;=\; \left(\frac{2\sqrt{y} - (-y)}{\sqrt{2}}\right)^2 \;=\; \frac{4y + 4y^{\frac{3}{2}} + y^2}{2}\)


\(\displaystyle \text{Therefore, the volume is: }\;V \;=\;\int^9_0\left(2y + 2y^{\frac{3}{2}} + \tfrac{1}{2}y^2\right)\,dy\)

\(\displaystyle \text{Got it?}\)

 

[BigGlenntheHeavy]

Good show soroban. Yes, I got it.