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

Explanation:

Step1: Find intersection points

Set $x^{2}-1 = 6$, then $x^{2}=7$, so $x=\sqrt{7}$ (since we are considering the region to the right of the $y -$axis, we take the positive root). The lower - bound of integration is $x = 0$ and the upper - bound is $x=\sqrt{7}$.

Step2: Determine the length of the leg of the isosceles right - triangle

The length of the leg $l$ of the isosceles right - triangle cross - section perpendicular to the $x$ - axis is given by the vertical distance between the line $y = 6$ and the curve $y=x^{2}-1$. So $l=6-(x^{2}-1)=7 - x^{2}$.

Step3: Find the area formula for the isosceles right - triangle

The area of an isosceles right - triangle with leg length $l$ is $A=\frac{1}{2}l^{2}$. Substituting $l = 7 - x^{2}$, we get $A(x)=\frac{1}{2}(7 - x^{2})^{2}=\frac{1}{2}(49-14x^{2}+x^{4})$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b=\sqrt{7}$, and $A(x)=\frac{1}{2}(49-14x^{2}+x^{4})$. So $V=\frac{1}{2}\int_{0}^{\sqrt{7}}(49-14x^{2}+x^{4})dx$. Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: [ \begin{align*} V&=\frac{1}{2}\left[49x-\frac{14}{3}x^{3}+\frac{1}{5}x^{5}\right]_{0}^{\sqrt{7}}\ &=\frac{1}{2}\left(49\sqrt{7}-\frac{14}{3}(\sqrt{7})^{3}+\frac{1}{5}(\sqrt{7})^{5}\right)\ &=\frac{1}{2}\sqrt{7}\left(49-\frac{14}{3}\times7+\frac{1}{5}\times49\right)\ &=\frac{\sqrt{7}}{2}\left(49-\frac{98}{3}+\frac{49}{5}\right)\ &=\frac{\sqrt{7}}{2}\left(\frac{735 - 490+147}{15}\right)\ &=\frac{\sqrt{7}}{2}\times\frac{392}{15}\ &\approx27.867 \end{align*} ]

Answer:

$27.867$