let r be the region bounded by the graphs of $f(x)=\\frac{6}{x+2}$ , $g(x)=4$ , and the vertical line…

Explanation:

Step1: Find y-range of region R

First, find the y-values at the bounds. At $x=1$, $f(1)=\frac{6}{1+2}=2$. The horizontal line is $g(x)=4$, so the region spans $y=2$ to $y=4$.

Step2: Express x as function of y

Rewrite $f(x)=\frac{6}{x+2}$ for $x$: $y=\frac{6}{x+2} \implies x+2=\frac{6}{y} \implies x=\frac{6}{y}-2$ The vertical line is $x=1$, so the length of the leg of the isosceles right triangle (the horizontal distance between $x=1$ and $x=\frac{6}{y}-2$) is: $L(y)=1 - \left(\frac{6}{y}-2\right)=3-\frac{6}{y}$

Step3: Area of cross-section

Area of an isosceles right triangle with leg length $L$ is $\frac{1}{2}L^2$. Substitute $L(y)$: $A(y)=\frac{1}{2}\left(3-\frac{6}{y}\right)^2=\frac{1}{2}\left(9-\frac{36}{y}+\frac{36}{y^2}\right)$

Step4: Integrate to find volume

Volume is the integral of the cross-sectional area over the y-interval $[2,4]$: $$ V=\int_{2}^{4} \frac{1}{2}\left(9-\frac{36}{y}+\frac{36}{y^2}\right) dy $$ First integrate term-by-term: $\int 9 dy = 9y$, $\int \frac{36}{y} dy = 36\ln|y|$, $\int \frac{36}{y^2} dy = -\frac{36}{y}$ So: $$ V=\frac{1}{2}\left[9y - 36\ln y - \frac{36}{y}\right]_{2}^{4} $$

Step5: Evaluate the definite integral

Calculate at upper bound $y=4$: $9(4)-36\ln4-\frac{36}{4}=36-36\ln4-9=27-36\ln4$ Calculate at lower bound $y=2$: $9(2)-36\ln2-\frac{36}{2}=18-36\ln2-18=-36\ln2$ Subtract lower from upper and multiply by $\frac{1}{2}$: $$ V=\frac{1}{2}\left[(27-36\ln4)-(-36\ln2)\right]=\frac{1}{2}\left[27-36(2\ln2)+36\ln2\right]=\frac{1}{2}\left[27-36\ln2\right] $$

Answer:

$\frac{27}{2}-18\ln2$ (or approximately $1.377$)