- #1
HeyHow!
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We have been given the task of finding the volume of a football (elliptical).
i know the area for an ellipse is [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]
where a=distance from center to major axis x-direction (Half the length of the ball) (a=14cm)
and
b=distance to minor axis y-direction (circumference/2pi) [tex]\frac{73}{2\pi}[/tex]
from there i get confused. i found on a website that
any help would be appreciated greatly. i am confused
i know the area for an ellipse is [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]
where a=distance from center to major axis x-direction (Half the length of the ball) (a=14cm)
and
b=distance to minor axis y-direction (circumference/2pi) [tex]\frac{73}{2\pi}[/tex]
from there i get confused. i found on a website that
i don't understand the Area*dx and where Area= pi*y^2 comes from. Does that mean that pi*y^2 has to be integrated? I also have no idea of how the integration would look like. i do not know what to integrate to get to (pi*b^2/a^2)* ((a^3) - ((a^3)/3)).Now if we integrate Area*dx where Area= pi*y^2 (area of a cross
section of the football as given for our ellipse above) between 0 and
5.5 we will obtain half the volume for our ideal football.
This integration results in:
(pi*b^2/a^2)* ((a^3) - ((a^3)/3))
One can simplify this equation into:
(2/3)*pi*a*(b^2)
Remember, this is half the volume of our ideal football. To be more
correct, one would integrate between -5.5 and 5.5. The calculations
work out easier using 0 to 5.5.
any help would be appreciated greatly. i am confused