Question

Please help me on this

Answer 1

$y+x=8\\ y=8-x\\\\ \dfrac{12}{x}=8-x\qquad x\not=0\\ 12=8x-x^2\\ x^2-8x+12=0\\ x^2-8x+16-4=0\\ (x-4)^2=4\\ x-4=2 \vee x-4=-2\\ x=6 \vee x=2\\\\ y=8-6 \vee y=8-2\\ y=2 \vee y=6\\ \boxed{(x,y)=\{(2,6),(6,2)\}}$

$\displaystyle S_P=\int\limits_2^68-x-\dfrac{12}{x}\, dx\\ S_P=\left[8x-\dfrac{x^2}{2}-12\ln x\right]_2^6\\ S_P=8\cdot6-\dfrac{6^2}{2}-12\ln 6-(8\cdot2-\dfrac{2^2}{2}-12\ln 2)\\ S_P=48-18-12\ln 6-16+2+12\ln2\\ S_P=16-12\ln 6+12\ln 2\\ S_P=16-12(\ln 6-\ln 2)\\\boxed{S_P=16-12\ln 3}$

$\displaystyle S_Q=\int\limits_2^6\dfrac{12}{x}\, dx\\ S_Q=\left[12\ln x\right]_2^6\\ S_Q=12\ln 6-12\ln 2\\ S_Q=12(\ln6-\ln 2)\\ \boxed{S_Q=12\ln 3}$

$\displaystyle V=\pi \int \limits_a^b f^2(x)\, dx\\\\ V=\pi \int \limits_2^6 \left(\dfrac{12}{x}\right)^2\, dx\\ V=\pi \int \limits_2^6 \dfrac{144}{x^2}\, dx\\ V=\pi\left[-\dfrac{144}{x}\right]_2^6\\ V=\pi \cdot \left(-\dfrac{144}{6}-\left(-\dfrac{144}{2}\right)\right)\\ V=\pi \cdot (-24+72)\\ V=\pi \cdot48\\ \boxed{V=48\pi}$