I suppose the ellipse should have the equation
[tex]\dfrac{x^2}{1600}+\dfrac{y^2}{400}=1\implies y=\pm\dfrac{\sqrt{1600-x^2}}4[/tex]
Then the cross sections are squares with side lengths
[tex]\dfrac{\sqrt{1600-x^2}}4-\left(-\dfrac{\sqrt{1600-x^2}}4\right)=\dfrac{\sqrt{1600-x^2}}2[/tex]
so that each section contributes an area of
[tex]\left(\dfrac{\sqrt{1600-x^2}}2\right)^2=\dfrac{1600-x^2}4[/tex]
Then the total volume of the pool would be given by
[tex]\displaystyle\int_{-40}^{40}\frac{1600-x^2}4\,\mathrm dx=\boxed{\frac{64,000}3}[/tex]
