Question

Consider a solid whose base is the ellipse 1x^2 + 25y^2 = 25 and whose cross-section perpendicular to the x-axis are square. What is the volume of this solid?

Answer 1

Notice the picture below, with an squared area, running along the ellipse, perpendicular to the x-axis

so $\bf x^2+25y^2=25\implies y^2=\cfrac{25-x^2}{25}\implies y=\sqrt{1-\cfrac{x^2}{25}} \\\\\\ \textit{that will be half of one side, notice in the picture}\\ \textit{so, a full side will be }2\sqrt{1-\cfrac{x^2}{25}} \\\\\\ \textit{now, the area of a square is }length\times width \\\\\\ but\quad \begin{cases} length=2\sqrt{1-\cfrac{x^2}{25}}\\\\ width=2\sqrt{1-\cfrac{x^2}{25}} \end{cases} \\\\\\ \textit{thus the area of it is }\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)$

$\bf or\quad \left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2$

now... how far is it from the left-side to the right-side of the ellipse?
well, that's the major axis of it
let's see which one is that in this case

$\bf x^2+25y^2=25\implies\cfrac{x^2}{25}+\cfrac{25y^2}{25}=1 \\\\\\ \cfrac{(x-0)^2}{5^2}+\cfrac{(y-0)^2}{1^2}=1\\\\ -----------------------------\\\\ \cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1 \ \ center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})\\\\ -----------------------------\\\\ \begin{cases} a=5\\ b=1\\ h=0\\ k=0 \end{cases}\implies vertices\to (0\pm 5,0)\to (\pm 5,0)$

so the ellipse runs from -5 to 5 over the x-axis

thus the area will be $\bf \displaystyle \int\limits_{-5}^{5}\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2 \\\\\\ \textit{or using the symmetrical property of the ellipse} \\\\\\ \displaystyle 2\int\limits_{0}^{5}\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2$