find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 25x² +…

Explanation:

Step1: Rewrite the ellipse equation

Rewrite $25x^{2}+16y^{2}=400$ as $y = \pm\frac{1}{4}\sqrt{400 - 25x^{2}}$. The length of the hypotenuse $h$ of the isosceles - right triangle cross - section perpendicular to the $x$ - axis is $h = y_{top}-y_{bottom}=\frac{1}{2}\sqrt{400 - 25x^{2}}$.

Step2: Find the side length of the isosceles - right triangle

For an isosceles right triangle with hypotenuse $h$, if the side length is $s$, then by the Pythagorean theorem $h^{2}=2s^{2}$, so $s=\frac{h}{\sqrt{2}}$. Here, $s=\frac{1}{2\sqrt{2}}\sqrt{400 - 25x^{2}}$.

Step3: Find the area of the cross - section

The area of an isosceles right triangle $A=\frac{1}{2}s^{2}$. Substituting $s$ into the formula, we get $A(x)=\frac{1}{2}\times(\frac{1}{2\sqrt{2}}\sqrt{400 - 25x^{2}})^{2}=\frac{400 - 25x^{2}}{16}$.

Step4: Determine the limits of integration

From the ellipse equation $25x^{2}+16y^{2}=400$, when $y = 0$, $x=\pm4$. So the limits of integration are $a=-4$ and $b = 4$.

Step5: Calculate the volume using the integral

The volume $V=\int_{a}^{b}A(x)dx=\int_{-4}^{4}\frac{400 - 25x^{2}}{16}dx$. Since the integrand is an even function, $V = 2\int_{0}^{4}\frac{400 - 25x^{2}}{16}dx$. [ \begin{align*} V&=2\times\frac{1}{16}\int_{0}^{4}(400 - 25x^{2})dx\ &=\frac{1}{8}(400x-\frac{25}{3}x^{3})\big|_{0}^{4}\ &=\frac{1}{8}(400\times4-\frac{25}{3}\times4^{3})\ &=\frac{1}{8}(1600-\frac{1600}{3})\ &=\frac{1}{8}\times\frac{3200}{3}\ &=\frac{400}{3} \end{align*} ]

Answer:

$\frac{400}{3}$