Hey guys, I need a bit of help with this question. I understand part i) but not part ii). I don't know how to figure it out and don't understand the solutions either Any help appreciated!

PS. please ignore the scribbles down the side of the page!

This will be added to the compilation. This is still NESA's solution but with more added depth (no pun intended)

\[ \textbf{2010 HSC Mathematics - Q10 b) ii)} \]

\[ \text{In part (1), a subtle thing to note is how they mention that the}\\ \text{hemispherical container is initially laid out }\textbf{horizontal.}\\ \text{That, and it is }\textbf{full.} \]

\[ \text{So initially it must've looked something like this.} \]

For reference I intentionally plot the centre of the circle.

\[ \text{Hence the initial depth is literally just the distance between the}\\ \text{centre of a circle, and any point on its circumference.}\\ \text{Which is of course, the radius, and hence equal to }r. \]

Note therefore that half of the initial depth must be \( \boxed{\frac{r}{2}} \).

\[ \text{Now after doing that rotation, the situation is as follows.}\\ \text{We wish to understand why.} \]

\[ \text{The }r\text{ towards the right should be reasonably clear.}\\ \text{However the red line segment }\textbf{also}\text{ has length }r. \]

\[ \text{This is because }\textbf{any}\text{ line from the centre of the circle to its circumference}\\ \text{has length }r\text{, where }r\text{ is the radius.}\\ \text{Note that the red line segment }\textbf{also}\text{ satisfies this.}\]

It just so happens that the red line is special, in that it's perpendicular to the ground (or whatever surface the container is lying on. And consequently the red line is vertical.

\[ \text{Again, the blue region shaded reflects the water in the bowl.}\\ \text{Note that the depth of the water is then just from the point where the bowl is at the ground}\\ \text{up to the black line, i.e. where the water goes up to.} \]

\[ \text{Since the question instructs us to make this depth equal to }\frac{r}{2}\\ \text{we do so. But then because the red line segment}\\ \text{has length }r\text{, everything above the water to the centre of the circle}\\ \text{must }\textbf{also}\text{ have length }\frac{r}{2}. \]

\[ \text{The last thing to justify is where the }\theta\text{ is placed.}\\ \text{This relies on the fact that }\theta\text{ measures simply what angle the bowl is }\textbf{tilted at}\\ \text{and therefore must be the angle drawn from the horizontal, as required.}\]

Or alternatively, if the diagram they gave was clear enough, you should be able to automatically see why \(\theta\) is there. I literally just copied \(\theta\) across - it was the \(r\) that required more deduction.

\[ \text{With the set-up done, the trigonometry is now easy.}\\ \text{Because we have opposite/hypotenuse here, we just consider}\\ \sin \theta = \frac{\frac{r}{2}}{r} \implies \sin \theta = \frac12 \implies \boxed{\theta = \frac\pi6}. \]

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\[ \text{Put simply, part (2) now wishes us to simplify the following fraction:}\\ \frac{\text{Remaining volume left}}{\text{Initial volume}} \]

Because we're given that the water initially fills the entire hemisphere, we just need to compute the volume of said hemisphere. Which is of course half that of the volume of the entire sphere, i.e. \(\frac12 \times \frac{4\pi r^3}{3}\), which simplifies to \( \boxed{\frac{2\pi r^3}{3}} \).

\[ \text{The remaining bit is to make use of part i).}\\ \text{It was more or less up to you to recognise that the 'solid' that the water manifests in}\\ \text{is basically the same as what you do in that volumes question above.} \]

By simply rotating NESA's own diagram by \(90^\circ\) anticlockwise this may be clearer. Focus on the grey shaded area, because that reflects the volume of the water.

Note that in the diagram in part i), only the top half is shaded. But recall that when you rotate to form your solid, you go through a full \(360^\circ\) rotation about the \(x\)-axis. By doing so, the bottom half of the minor segment does get included as well.

\[ \text{So in our case, the volume of the water is just the answer in part i)}\\ \text{except thanks to (1), we know now that }\theta = \frac\pi6.\]

\[ \text{Thus we get}\\ \begin{align*}V&= \frac{\pi r^3}{3} \left(2-3\sin \frac\pi6 +\sin^3\frac\pi6 \right)\\ &= \frac{\pi r^3}{3} \times \frac{5}{8}\\ &= \frac{5\pi r^3}{24}. \end{align*} \]

\[ \text{Finally subbing back, the required fraction is}\\ \frac{ \frac{5\pi r^3}{24}}{\frac{2\pi r^3}{3}} = \frac{5}{16}. \]