Create a bucket by rotating around the y axis the curve y=3ln(x−6) from y = 0 to y = 4. If this bucket contains a liquid with density 780 kg/m3 filled to a height of 3 meters, find the work required to pump the liquid out of this bucket (over the top edge). Use 9.8 m/s2 for gravity.

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A bucket formed by rotating aroune the y-akis\begin{array}{l}y=3 \text { en }(x-6) \quad \text { from } y=0 \text { to } y=4 \\\rho=780 \mathrm{~kg} / \mathrm{m}^{3} \quad y=9.8 \mathrm{~m} / \mathrm{sec} . \\\text { height }=3 \mathrm{~m} . \\y=0 \text { when } x=7, x=6+e^{4 / 3} \Rightarrow x=6+e^{y / 3}\end{array}work wr fol =m g \times dws \int \rho d g d\begin{array}{l}w=\int_{0}^{3} \rho \pi\left(6+e^{y / 3}\right)^{2} d y \cdot g \cdot(4-y) \\w=\rho g \pi \int_{0}^{3}\left(36+e^{2 y / 3}+12 e^{y / 3}\right)(4-y) d y \\=99 \pi\left[\int_{0}^{3} 144 d y-\int_{0}^{3} 36 y d y+\int_{0}^{3} 4 e^{2 y / 3} d y\right. \\\begin{array}{c}-\int_{0}^{3} y e^{2 y / 3} d y+48 \int_{0}^{3 y / 3} e^{y / 3} d y-1 / x+x \\\left.-\int_{0}^{3} 2 y e^{y / 3} d y\right]\end{array} \\=99 \pi\left[432-162+6\left(e^{2}-1\right)-\left(\frac{6 y-9}{4} \cdot e^{\frac{2 y}{3}}\right)_{0}^{3}+144(e-1)\right. \\-108] \\\end{array}so\begin{aligned}w & =99 \pi\left[162+6 e^{2}-6+144 e-144-\frac{9 e^{2}}{4}-\frac{9}{4}\right] \\& =780 \times 9.8 \times \pi\left[12-\frac{9}{4}+6 e^{2}+144 e-\frac{9 e^{2}}{4}\right] \\& =10299544 \cdot 9 \text { Jovle. }\end{aligned}sir I rounded to one decimal. If any mistake plz comment  ...