Tilted cylindrical tank - 2

Align the axis of the cylinder along the $x$ -axis with the water surface of the tilted cylinder be symmetrically divided by the $xz$ -plane. We note that the cross-sectional area perpendicular to the $x$ -axis at $x$ is $A = \pi r^2 - r^2 \cos^{-1} \left(\dfrac hr \right) + h \sqrt{r^2 - h^2}$ , where $r=3$ and $h(x)$ is a function of $x$ (the blue line). Let the point on the water surface farthest from the $z$ -axis be $b$ . Then $\dfrac {h(x)+3}{x-b} = - \sqrt 3 \implies h(x) = \sqrt 3(b-x) - 3$ . As the volume of the water in the tank is $9\pi$ , we have:

$9\pi = \int_0^b \left(9\pi - 9\cos^{-1} \left(\frac h3 \right) + h\sqrt{9-h^2}\right) dx = 9\pi b - 9 \int_0^b \cos^{-1} \left(\frac h3 \right) dx + \int_0^b h\sqrt{9-h^2}\ dx$

Consider the last two integrals separately,

$\begin{aligned} I & = \int_0^b \cos \frac h3 \ dx & \small \blue{\text{Let }u = \frac h3} \\ & = \sqrt 3 \int_{-1}^{\frac b{\sqrt 3}-1} \cos u \ du & \small \blue{\implies du = \frac {dh}3 = - \frac {dx}{\sqrt 3}} \\ & = \sqrt 3 \left[u \cos^{-1} u + \int \frac u{\sqrt{1-u^2}} du \right]_{-1}^{\frac b{\sqrt 3}-1} & \small \blue{\text{Integration by parts}} \\ & = \sqrt 3 \left[u \cos^{-1} u - \sqrt{1-u^2} \right]_{-1}^{\frac b{\sqrt 3}-1} \\ & = \sqrt 3 \left[\left(\frac b{\sqrt 3}-1\right) \cos^{-1} \left(\frac b{\sqrt 3}-1\right) - \sqrt{\frac {2b}{\sqrt 3}-\frac {b^2}3} + \pi\right] \\ & = (b - \sqrt 3) \cos^{-1} \left(\frac b{\sqrt 3}-1\right) - \sqrt{2\sqrt 3b - b^2} + \sqrt 3 \pi \end{aligned}$

$\begin{aligned} J & = \int_0^b h \sqrt{9-h^2} \ dx & \small \blue{\text{Let }\cos \theta = \frac h3} \\ & = 9\sqrt 3 \int_{\cos^{-1}\left(\frac b{\sqrt 3}-1\right)}^\pi \sin^2 \theta \cos \theta \ d\theta & \small \blue{\implies - \sin \theta \ d\theta = \frac {dh}3 = - \frac {dx}{\sqrt 3}} \\ & = 9\sqrt 3 \int_{\sqrt{\frac {2\sqrt 3 b-b^2}3}}^0 v^2 \ dv \\ & = - \left(2\sqrt 3 b - b^2\right)^\frac 32 \end{aligned}$

Therefore we have:

$9\pi b - 9(b - \sqrt 3) \cos^{-1} \left(\frac b{\sqrt 3}-1\right) - 9\sqrt{2\sqrt 3b - b^2} + 9\sqrt 3 \pi - \left(2\sqrt 3 b - b^2\right)^\frac 32 = 9\pi$

Solving it numerically, we get $b \approx 2.67709164$ . And the height from the ground of the water surface $h = b \cos 30^\circ \approx \boxed{2.318}$ .