shuning
Member
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- Aug 23, 2008
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- HSC
- 2009
the area enclosed between the parabola
and its latus rectum is ![](https://latex.codecogs.com/png.latex?\bg_white \frac{8a^2}{3} units^3)
a solid figure has its base in the xy plane, the eclipse [text]\frac{x^2}{16}+\frac{y^2}{4}=1[/tex]
cross-sections perpendicular to the x-axis are parabolas with latus rectums in the xy plane
show that the area of the cross section at x=h is![](https://latex.codecogs.com/png.latex?\bg_white \frac{16-h^2}{6} units^3)
my working:
![](https://latex.codecogs.com/png.latex?\bg_white A =\frac{8a^2}{3} )
![](https://latex.codecogs.com/png.latex?\bg_white A = \frac{8y^2}{3} )
![](https://latex.codecogs.com/png.latex?\bg_white A = \frac{8(4 - \frac{h^2}{4})}{3} )
![](https://latex.codecogs.com/png.latex?\bg_white A = \frac{32-2h^2}{3} )
WHERE THE FUCK WAS I WRONG????
a solid figure has its base in the xy plane, the eclipse [text]\frac{x^2}{16}+\frac{y^2}{4}=1[/tex]
cross-sections perpendicular to the x-axis are parabolas with latus rectums in the xy plane
show that the area of the cross section at x=h is
my working:
WHERE THE FUCK WAS I WRONG????