(source: gyazo.com)
I need help for part $(i)$ ... What I think I know so far:
$P = dgh = (1000)(9.8)(h)$
Finding $h$:
- We will choose an arbitrary value '$x$' somewhere between $0$ and $3$.
- The height of this "plane" is $3-x$.
- Unfortunately, the volume is difficult to calculate because $r$ is a function of $x$.
- We will need to create a formula that relates $r$ as $x$ increases. This is possibly modeled by $r(x)=\sqrt{25-x^2}$
What am I missing here ?
Hint: The $3$ is really not relevant at this stage, it will only come into play when you are choosing the limits of integration.
Draw a vertical cross-section that goes through the centre $C$ of the hemisphere. You have probably already done this. It is essential.
The distance from $C$ to the centre of the circle of radius $r$ we are curious about is $5-x$. The distance from $C$ to the edge of that circle is $5$. So by the Pythagorean Theorem, we have $(5-x)^2+r^2=5^2$. That yields $r^2=10x-x^2$. So the area of our circle of radius $r$ is $\pi(10x-x^2)$.
That answers your question about $r$. We will need to lift that slice through a distance $6-x$, so your integrand will ultimately be a constant times $(6-x)(10x-x^2)$.