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.)

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 = $\displaystyle 2x$

area of the square = $\displaystyle 4x^2$

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

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

finish it

Re: question on the volume of a pool in the shape of an elipse

So for a given y, the two x values are $\displaystyle x= \pm 5\sqrt{1- y^2/16}= \pm\frac{5}{4}\sqrt{16- y^2}$ and the length of that segment is $\displaystyle \frac{5}{2}\sqrt{16- y^2}$. Since "cross sections perpendicular to the ground and parallel to the y-axis are squares", the depth at that y is also $\displaystyle \frac{5}{2}\sqrt{16- y^2}$ and so the cross section area is the square of that, $\displaystyle \frac{25}{4}(16- y^2)$. Taking the thickness of an "infinitesmal slab" to be [itex]dy[/tex], the volume of such a slab is [itex]\frac{25}{4}(16- y^2)dy[/itex] and the total volume is the integral of that.