This question aims to find the volume of the pool having square cross-sections perpendicular to the $x-$axis.

The numerical measure of a three-dimensional solid is said to be the volume. The concept is similar to the measurement of the area in a two-dimensional region. Usually, some of the fundamental geometric concepts are used to calculate the volume of a solid. When a plane intersects with a solid, it forms the cross-section area. There will be no particular formula for the volume of a solid having an inconstant cross-section. In such a case, the volume can be calculated with the aid of a definite integral. This can be accomplished by following the steps: first, slice the solid into pieces, second calculate the volume of each slice and sum up all. The condition to get the volume of the whole solid is that all the slices must be parallel to one another.

## Expert Answer

Assume the origin to be at the pool’s center. Also, each strip perpendicular to the $x-$axis is shaped like a square. Since the origin is at the center of the pool, therefore, the $y-$coordinate equals $\dfrac{1}{2}$ of the pool’s width at that location for every $x$ value. Now, for the square cross-sections:

Width$=$Depth $=2y$

Cross-section area $=(2y)(2y)=4y^2$

Let $dx$ represent the thickness of every slice then volume of every cross-section will be:

$dV=A(x)\,dx=4y^2\,dx$

Equation of the ellipse is given by:

$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}=1$

So that:

$y^2=b^2\left(1-\dfrac{x^2}{a^2}\right)$

and $dV=4b^2\left(1-\dfrac{x^2}{a^2}\right)\,dx$

Now we compute the volume by integrating $dV$. Since the pool is symmetric about the $x-$axis so, first compute the volume of the half pool and multiply it by $2$ to obtain the total volume. This makes the integration limits from $x=0$ to the pool’s edge which is at $x=a$.

$V_h=\int dV=\int\limits_{0}^{a}4b^2\left(1-\dfrac{x^2}{a^2}\right)\,dx$

$=4b^2\int\limits_{0}^{a}\left(1-\dfrac{x^2}{a^2}\right)\,dx$

$=4b^2\left|x-\dfrac{x^3}{3a^2}\right|_{0}^{a}$

$=4b^2\left[a-\dfrac{a^3}{3a^2}-0\right]$

$=4b^2\left[a-\dfrac{a}{3}\right]$

$=\dfrac{8ab^2}{3}$

And so the total volume is given by:

$V=2\times \dfrac{8ab^2}{3}$

$V=\dfrac{16ab^2}{3}$

Now from the given equation, and the equation of the ellipse, we have:

$a^2=2500\implies a=50$Â and $b^2=900$

Therefore, $V=\dfrac{16(50)(900)}{3}$

$V=240,000\,ft^3$

## Example

The radius and height of the cone are $3$ and $2$, respectively. Find its volume.

### Solution

Here, $r=3$ and $h=2$.

Let the vertex of the cone be at the origin and the axis of the cone be at $x-$axis. The radius of the cross-section at $x-$axis is:

$y=\dfrac{rx}{h}=\dfrac{3}{2}x$

And $A(x)=\dfrac{\pi r^2 x^2}{h^2}=\dfrac{9\pi}{4}x^2$

So, $V=\int\limits_{0}^{2}\dfrac{9\pi}{4}x^2\,dx$

$V=\dfrac{9\pi}{4}\int\limits_{0}^{2}x^2\,dx$

$V=\dfrac{9\pi}{4}\left|\dfrac{x^3}{3}\right|_{0}^{2}$

$V=\dfrac{9\pi}{4}\left[\dfrac{8}{3}-0\right]$

$V=6\pi$