question let the region r be the area enclosed by the function f(x)=x² - 1, the horizontal line y = 6, and…

Explanation:

Step1: Find intersection points

Set $x^{2}-1 = 6$, then $x^{2}=7$, so $x=\pm\sqrt{7}$.

Step2: Determine the length of the leg of the triangle

The length of the leg of the isosceles - right - triangle cross - section perpendicular to the $x$ - axis is $l = 6-(x^{2}-1)=7 - x^{2}$.

Step3: Find the area of the cross - section

The area formula for an isosceles right - triangle is $A=\frac{1}{2}s^{2}$, where $s$ is the length of a leg. So $A(x)=\frac{1}{2}(7 - x^{2})^{2}=\frac{1}{2}(49-14x^{2}+x^{4})$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x =-\sqrt{7}$ to $x=\sqrt{7}$ is given by $V=\int_{-\sqrt{7}}^{\sqrt{7}}A(x)dx$. Since $A(x)$ is an even function, $V = 2\int_{0}^{\sqrt{7}}\frac{1}{2}(49-14x^{2}+x^{4})dx=\int_{0}^{\sqrt{7}}(49-14x^{2}+x^{4})dx$. Integrating term - by - term: $\int(49-14x^{2}+x^{4})dx=49x-\frac{14}{3}x^{3}+\frac{1}{5}x^{5}+C$. Evaluating the definite integral: $\left[49x-\frac{14}{3}x^{3}+\frac{1}{5}x^{5}\right]_{0}^{\sqrt{7}}=49\sqrt{7}-\frac{14}{3}(\sqrt{7})^{3}+\frac{1}{5}(\sqrt{7})^{5}$. $49\sqrt{7}-\frac{14}{3}\times7\sqrt{7}+\frac{1}{5}\times49\sqrt{7}=\sqrt{7}(49-\frac{98}{3}+\frac{49}{5})$. $49-\frac{98}{3}+\frac{49}{5}=\frac{49\times15 - 98\times5+49\times3}{15}=\frac{735 - 490 + 147}{15}=\frac{392}{15}$. So $V=\frac{392\sqrt{7}}{15}\approx68.297$.

Answer:

$68.297$