# surface of revolutions on a full vessel

I'm trying to solve this question. All I can think of is the sruface of revolutions, and got the xbounds, but am really stuck on the setup. Thanks in advance.

Work definition I know is mgsintheta, but am also not sure on how to put in the integration

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Volume: Let $A(y)$ be the area of cross-section at "height" $y$. The cross-section is a circle of radius $x$, where $y=2(x^2-4)$. Note that when $x=0$ we have $y=-8$.
The area of cross-section is $\pi x^2$. Express this in terms of $y$. From $y=2(x^2-4)$ we find that $x^2=\frac{y}{2}+4$. So we want $$\int_{-8}^0\pi\left(\frac{y}{2}+4\right)\,dy.$$ I think we get $16\pi$.
Work: Take a slice of "thickness" $dy$ at level $y$, where $y$ goes from $-8$ to $0$. We want to lift this slice to height $y=4$, so through a distance $4-y$. (Note that $y$ will be negative throughout, so $4-y\gt 4$.)
The volume of the slice is the area of cross-section times $dy$. By our previous work, the area of cross-section is $\pi\left(\frac{y}{2}+4\right)$. So the total work done is $$\int_{-8}^0 K(4-y)\pi\left(\frac{y}{2}+4\right)\,dy,$$ where $K$ is a constant that depends on the units used, and the value one uses for $g$.