clac 2

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\begin{array}{l}y=4 \ln (x-3) \\ \frac{y}{4}=\ln (x-3) \\ x-3=e^{\frac{y}{4}} \\ x=e^{\frac{y}{4}}+3 \\ \text { Work done } W=F \cdot d \\ =m g \cdot d \\ =\rho v g \cdot d \\ =\rho\left(\pi x^{2} d y\right) g(8-y) \\ =\pi \rho g\left(e^{\left.\frac{y}{4}+3\right)^{2}(8-y) d y}\right. \\ w=\int_{0}^{3} \pi \rho g\left(e^{\frac{y}{2}}+6 e^{\frac{y}{4}}+9\right)(8-y) d y \\ =\pi \rho g \int_{0}^{3}\left(8 e^{\frac{y}{2}}-e^{\frac{y}{2}} y+48 e^{\frac{y}{4}}-6 e^{\frac{y}{4}} y+72-9 y\right) d y \\ =\pi \rho g\left[16\left(e^{\frac{3}{2}-1}\right)-4\left(\frac{e^{3 / 2}}{2}+1\right)+192\left(e^{3 / 4}-1\right)-6\left(-4 e^{3 / 4}+16\right)+216-\frac{8}{2}\right] \\ =3.141 * 740 * 9.8\left[14 e^{3 / 2}+216 e^{3 / 4}-\frac{265}{2}\right] \\ W=8827037.65 \mathrm{~J} \\\end{array} ...