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

Explanation:

Step1: Find intersection points

Set $x^{2}=6$, so $x = \sqrt{6}$ (since we are in the first - quadrant as we are bounded by the $y$-axis).

Step2: Determine the area formula for the cross - section

The area of an isosceles right - triangle with leg length $l$ is $A=\frac{1}{2}l^{2}$. Here, the leg length $l$ of the isosceles right - triangle cross - section perpendicular to the $x$-axis is $6 - x^{2}$. So, $A(x)=\frac{1}{2}(6 - x^{2})^{2}=\frac{1}{2}(36-12x^{2}+x^{4})$.

Step3: Use the volume formula for cross - sectional area

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = 0$ to $x=\sqrt{6}$ is given by the integral $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b=\sqrt{6}$, and $A(x)=\frac{1}{2}(36-12x^{2}+x^{4})$. So, $V=\int_{0}^{\sqrt{6}}\frac{1}{2}(36 - 12x^{2}+x^{4})dx$.

Step4: Integrate term - by - term

$\int_{0}^{\sqrt{6}}\frac{1}{2}(36 - 12x^{2}+x^{4})dx=\frac{1}{2}\int_{0}^{\sqrt{6}}(36 - 12x^{2}+x^{4})dx=\frac{1}{2}\left[36x-4x^{3}+\frac{1}{5}x^{5}\right]_{0}^{\sqrt{6}}$.

Step5: Evaluate the definite integral

$\frac{1}{2}\left(36\sqrt{6}-4(\sqrt{6})^{3}+\frac{1}{5}(\sqrt{6})^{5}\right)=\frac{1}{2}\left(36\sqrt{6}-4\times6\sqrt{6}+\frac{1}{5}\times6^{2}\sqrt{6}\right)=\frac{1}{2}\sqrt{6}\left(36 - 24+\frac{36}{5}\right)=\frac{1}{2}\sqrt{6}\left(12+\frac{36}{5}\right)=\frac{1}{2}\sqrt{6}\times\frac{60 + 36}{5}=\frac{1}{2}\sqrt{6}\times\frac{96}{5}=\frac{48\sqrt{6}}{5}\approx23.516$.

Answer:

$23.516$