This might be a somewhat unorthodox question. I am wondering if there are any simple guidelines or tips/tricks to better understand volumes of solids of revolution.
E.g. the simple assignment and my own thoughts. Please correct me!
$\text{Let a and h be positive, let A be the area in the first quadrant bounded by }$$ax^2$$ \text{the y-axis and y=h.}$
The area $A$ is rotated about the $y$-axis. Find the volume of the solid created.
My initial thoughts would be to graph a rough sketch of the scenario. So I have a shape bounded by $ax^2$ on the right, by $x=0$ on the left, and $y=h$ at the top. The lower bound is not relevant, as it is 0 before it crosses the axis.
So, if I set $y=h \land y=ax^2$ to be equal, I can find the $x$-value where I need to integrate to, i.e. $ax^2=h \implies x=\sqrt{\dfrac{h}{a}}$.
My guess would be: $$\pi\int_0^{\sqrt{\tfrac{h}{a}}}(ax^2)^2 - (h)^2 dx \implies$$
$$\pi\int_0^{\sqrt{\tfrac{h}{a}}}(a^2x^4) - \dfrac{a^2h^2}{a^2}\,dx=\pi a^2\int_0^{\sqrt{\tfrac{h}{a}}}x^4-\dfrac{h^2}{a^2}\,dx = \pi a^2 \left( \int_0^{\sqrt{\tfrac{h}{a}}}x^4\,dx - \dfrac{h^2}{a^2}{\sqrt{\dfrac{h}{a}}}\right)$$
$$\pi a^2 \left(\dfrac{\sqrt{\tfrac{h}{a}}^5}{5} - \dfrac{h^2}{a^2}{\sqrt{\dfrac{h}{a}}}\right) = \dots$$
So, at this point I am both algebraically stuck and somewhat close to giving up. Any advice?
Cheers.
Hint: if a volume is defined by "stacked" plane surfaces of area $A(y)$ for $y$ from $0$ to $H$, then the volume is given by
$$V=\int_0^H A(y) \, \mathrm{d}y$$
Here, your plane surfaces are discs, whose radii are defined by the $y=ax^2$ curve.
That is, $R(y) = x=\sqrt{y/a}$, so the area is $A(y)=\pi R(y)^2=\pi y/a$, and
$$V=\int_0^{h} \pi \frac y a \mathrm{d}y$$
Can you conclude now?