You have a couple of errors: as you suspected, the height isn’t quite right, but the cross-sectional area is also wrong.
The radius of the tank at height $y$ is $\frac25y$, so the cross-sectional area is $\frac{4\pi}{25}y^2$. To raise the cross-sectional slice of thickness $dy$ at height $y$ to the top of the tank, you must raise it $15-y$ feet, doing $$dW=64\cdot\frac{4\pi}{25}y^2(15-y)dy=\frac{256\pi}{25}(15y^2-y^3)dy$$ foot-pounds of work. Thus, it takes
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy$$
foot-pounds of work to raise all of the water to the top of the tank.
There are $64\cdot\frac13\cdot16\pi\cdot10=\frac{10240\pi}3$ pounds of water in the tank; they all fall a distance of $115-18=97$ feet after being raised to the top of the tank, doing $97\cdot\frac{10240\pi}3=\frac{993,280\pi}3$ foot-pounds of work in the process. Thus, the net work required is
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy-\frac{993,280\pi}3\text{ foot-pounds}\;.$$
You can of course set it up as a single integral: the slice at height $y$ is first raised $15-y$ feet and then dropped $97$ feet, so it is ‘raised’ a total of $15-y-97=-y-82$; your error was getting the wrong algebraic sign on $y$. If you do this, the answer is
$$\frac{256\pi}{25}\int_0^{10}(-y^3-82y^2)dy$$
foot-pounds.
Assume that $r$= radius of cylinder = radius of cone = height of cylinder = height of cone.
Hence we can calulate the following:
- Volume of cylinder = volume of water = $\pi r^3$
- Volume of cone = $\frac13 \pi r^3$
- Volume of container = total volumes of cylinder and conee = $\frac 43 \pi r^3$
- Volume of air in container = $\frac 13 \pi r^3$
We know that
- (from above) the volume of water is more than half the volume of container
- container is symmetrical about its longitudinal axis
hence when container is lying on its side, the level of water must be greater than the midheight, $r$, of the container (on its side).
Let the level of water in the container on its side be $r+a$ where $0<a<1$.
(a) Cylinder
Volume of air is the area of a circular segment multiplied by its height $r$, i.e.
$$\begin{align}
V_1&=\frac 12 r^3 \ (2\alpha - \sin 2\alpha)\cdot r\qquad
\text{where $\cos\alpha=\frac ar$}\\
&=\frac 12 r^3\ (2\alpha-2\sin\alpha\cos\alpha)\\
&=r^3\left[\cos^{-1}\left(\frac ar\right) -\left(\frac ar\right)\sqrt{1-\left(\frac ar\right)^2}\ \ \right]\\
&=r^3(\cos^{-1}\lambda-\lambda\sqrt{1-\lambda^2})\qquad
\text{where $\lambda=\frac ar$}\\
&=\frac{r^3}3\left[3\cos^{-1}\lambda-3\lambda\sqrt{1-\lambda^2}\right]
\end{align}$$
(b) Cone
Calculate the the volume of air in the cone by integrating circle segments of infinitesimally small thickness and of different radius $u$, as $u$ varies from $r$ (the part adjacent to the cylinder) to $a$ (edge of container).
Area of circle segment with radius $u$ is given by
$$\frac 12 u^2 \ (2\phi - \sin 2\phi)
= u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2} \;\right]\qquad
\text{where $\cos\phi=\frac au$}\\$$
Volume of air in cone is given by integrating this from $u=r$ to $u=a$, i.e.
$$\begin{align}
V_2&=\int_r^a
u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2}\;\right] du\\
&=\int_r^a u^2\cos^{-1} \left(\frac au\right) -a\sqrt{u^2-a^2} \; du\\
&=a^2 \int_r^a \left(\frac ua \right)^2\cos^{-1} \left(\frac au\right) -\sqrt{\left(\frac ua\right)^2-1} \; \; du\
\end{align}$$
Put
$$\begin{align}
\frac au = \cos\phi \Rightarrow
u&=a\sec\phi\quad \Rightarrow \sec\phi=\frac ua\\
du&=a\sec\phi\tan\phi \; d\phi\\
u=r \Rightarrow \phi&=\cos^{-1}\frac ar=\cos^{-1}\lambda\\
u=a \Rightarrow \phi&=\cos^{-1}1=0
\end{align}$$
This gives
$$\begin{align}
V_2&=a^2\int_{\cos^{-1}\lambda}^{0}\left[(\sec^2\phi) \cdot \phi-\sqrt{\sec^2\phi-1}\right]\ a\sec\phi\tan\phi \; d\phi\\
&=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec\phi\tan^2\phi \; d\phi\\
&=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec^3\phi +\sec\phi \; d\phi\\
&=\cdots \text{(and after some tedious integration)}\cdots\\
&=\frac {a^3}3 \left[\phi\sec^3\phi-2\sec\phi\sqrt{\sec^2\phi-1}+\ln(\sec\phi+\ sqrt{\sec^2\phi-1})\right]_0^{\cos^{-1}\lambda}\\
&=\frac {a^3}3\left[ \left(\cos^{-1}\frac au\right)\left(\frac ua\right)^3-2\frac ua\sqrt{\left(\frac ua\right)^2-1}+\ln\left(\frac ua+\sqrt{\left(\frac ua\right)^2-1}\right)\right]_{u=a}^{u=r}\\
&=\frac {a^3}3 \left[ \left(\cos^{-1} \frac ar\right)\left(\frac ra\right)^3-2\frac ra \sqrt{\left(\frac ra\right)^2-1}+\ln\left(\frac ra+\sqrt{\left(\frac ra\right)^2-1}\right)\right]\\
&=\frac {a^3}3 \left[ \frac 1{\lambda^3} \cos^{-1}\lambda -\frac 2\lambda \sqrt{\frac 1{\lambda^2}-1}+\ln\left(\frac 1\lambda+\sqrt{\frac 1{\lambda^2}-1}\right) \right]\\
&=\frac {r^3}3 \left[ \cos^{-1} \lambda-2\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda +\sqrt{\frac 1{\lambda^2}-1}\right)\right]
\end{align}$$
(c) Total
Total volume of air in container is
$$\begin{align}
V_1+V_2&=\frac {\pi r^3}3\\
\frac{r^3}3\left[4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)\right]&=\frac {\pi r^3}3\\
4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)-\pi&=0\\
\lambda &\approx 0.36876\\
a=\lambda r &\approx 0.36876 r
\end{align}$$
Hence, height of water in container on its side is
$$r+a=(1+\lambda)r\approx 1.36876r \qquad \blacksquare$$
NB:
1. See this for a useful guide to the messy integration and the volume of a vertical cut of a cone.
2. See this for a useful guide for integrating $\sec^3\theta$.
Best Answer
The tank is a truncated circular cone with a base of radius $r=3\,$m, a depth of $4\,$m and top radius of $r=4\,$m. Let $y$ denote the vertical distance from the bottom of the tank. Then $r=3+\frac{1}{4}y$ is the radius of the slice of water lying $y$ meters above the bottom of the tank.
Think of that slice of water as being solid rather than liquid, as if it were frozen. The amount of work done in pumping the water to a height of $5\,$m above the base is the same as if one had to move all those frozen slices to that height.
The volume of each slice is $\pi r^2\,dy$ with $dy$ representing the thickness. The mass of the slice (in the mks system) is found by multiplying the density $\rho=1000\,$ kg/m$^3$ times the volume, so \begin{equation} M=1000\pi r^2\,dy=1000\pi\left(3+\frac{1}{4}y\right)^2dy \end{equation}
Each of these slices of water must be moved upward a distance of $D=5-y$ meters against a gravitational force of $g=9.8\,$m/sec$^2$ resulting in the work for the slice at $y$ being
\begin{equation} W_y=9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy \end{equation}
The total work to remove all the slices of water between $y=0$ and $y=4$ is
\begin{equation} W=\int_{0}^{4}9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy \end{equation}
The rest is routine since the integrand is simply a third degree polynomial. The units will be joules.