# volume of elliptical object

• Nov 29th 2008, 03:50 AM
mat
volume of elliptical object
hello I have to find the volume in metres cubed and in litres cubed of an above ground swimming pool. the surface area of the pool is of an axis 2 metres and 3 metres.the depth is 1.3 metres. so 2 times 3 times 1.3m=7.8 metres cubed. how to find how many litres correct to the nearest 10 litres? I need answer in about 24 hours thanks.
• Nov 29th 2008, 04:45 AM
earboth
Quote:

Originally Posted by mat
hello I have to find the volume in metres cubed and in litres cubed of an above ground swimming pool. the surface area of the pool is of an axis 2 metres and 3 metres.the depth is 1.3 metres. so 2 times 3 times 1.3m=7.8 metres cubed. how to find how many litres correct to the nearest 10 litres? I need answer in about 24 hours thanks.

According to your headline and the wording of the question I assume that the base area of the pool has to be an ellipse.

1. Area of an ellipse:

$A_{elli} = \pi \cdot a \cdot b$ ...... Sub in the values for a and b (in meters!)

$A_{elli}=\pi \cdot 2 \cdot 3 = 6\pi\ m^2$ ...... therefore the volume of the pool is:

$V_{pool}=A_{elli} \cdot 1.3\ m = 7.8\pi\ m^3\approx24.5044227\ m^3$

Use the property: One liter is the same as one decimeter cubed and one decimeter is the same as 10 cm. Therefore: $1m^3 = 1,000\ l$

So the pool contains:

$V_{pool}=24.5044227\ m^3 \cdot 1,000\frac{l}{m^3}= 24504.4227\ dm^3 \approx 24500\ l$ to the nearest 10 l.