# question on the volume of a pool in the shape of an elipse

• May 9th 2012, 01:21 PM
Michellemo
question on the volume of a pool in the shape of an elipse
As viewed from above, a swimming pool has the shape of the ellipse
x^2/25 +y^2/16=1
The cross sections perpendicular to the ground and parallel to the y-axis are squares.
Find the total volume of the pool. (Assume the units of length and area are meters
and square meters respectively.)
• May 9th 2012, 01:53 PM
skeeter
Re: question on the volume of a pool in the shape of an elipse
Quote:

Originally Posted by Michellemo
As viewed from above, a swimming pool has the shape of the ellipse
x^2/25 +y^2/16=1
The cross sections perpendicular to the ground and parallel to the y-axis are squares.
Find the total volume of the pool. (Assume the units of length and area are meters
and square meters respectively.)

side of a square = $2x$

area of the square = $4x^2$

$x^2 = 25\left(1 - \frac{y^2}{16}\right)$

$V = 2\int_0^4 A(y) \, dy$

finish it
• May 9th 2012, 01:53 PM
HallsofIvy
Re: question on the volume of a pool in the shape of an elipse
So for a given y, the two x values are $x= \pm 5\sqrt{1- y^2/16}= \pm\frac{5}{4}\sqrt{16- y^2}$ and the length of that segment is $\frac{5}{2}\sqrt{16- y^2}$. Since "cross sections perpendicular to the ground and parallel to the y-axis are squares", the depth at that y is also $\frac{5}{2}\sqrt{16- y^2}$ and so the cross section area is the square of that, $\frac{25}{4}(16- y^2)$. Taking the thickness of an "infinitesmal slab" to be $dy[/tex], the volume of such a slab is [itex]\frac{25}{4}(16- y^2)dy$ and the total volume is the integral of that.