shuning
Member
- Joined
- Aug 23, 2008
- Messages
- 654
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- Male
- HSC
- 2009
the area enclosed between the parabola 
and its latus rectum is 
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:
}{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????
