Q.Find, by integration, the volume of the solid generated by revolving about the $y$-axis, the region enclosed by $x^2=1+y$ and $y=3$.

The disc method $V=\pi\int y^2\,dx$ applied directly to $y^2=x(a-x)$ over its natural bounds $[0,a]$ gives $\pi a^3/6$.

1. The curve $y^2=x(a-x)$ meets the $x$-axis where $y=0$: $x(a-x)=0\Rightarrow x=0$ or $x=a$, so the bounded region runs from $x=0$ to $x=a$.
2. Volume of revolution about the $x$-axis (disc method): $V=\pi\displaystyle\int_0^ay^2\,dx=\pi\int_0^ax(a-x)\,dx$.
3. Expand: $\pi\displaystyle\int_0^a(ax-x^2)\,dx=\pi\left[\dfrac{ax^2}2-\dfrac{x^3}3\right]_0^a$. …

The triangle with vertices $(0,0),(3,0),(3,3)$ rotated about the $x$-axis forms a cone of radius 3 and height 3, whose volume by both the cone formula and direct integration is $9\pi$.

1. The three sides of the triangle are: the segment on the $x$-axis from $(0,0)$ to $(3,0)$ (i.e. $y=0$); the vertical segment from $(3,0)$ to $(3,3)$ (i.e. $x=3$); and the segment from $(0,0)$ to $(3,3)$, which lies on the line $y=x$.
2. Rotating this triangular region about the $x$-axis sweeps out a solid whose radius at position $x$ is $y=x$ (the hypotenuse), for $x$ from $0$ to $3$ — this is exactly a right circular cone with apex at the origin, base radius $3$ (at $x=3$), and height $3$. …

Revolving the ellipse about its major axis gives volume $\frac43\pi ab^2$, and about its minor axis gives $\frac43\pi a^2b$; their ratio is $b:a$.

1. Ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \Rightarrow y^2=b^2\left(1-\dfrac{x^2}{a^2}\right)$.
2. Volume when revolved about the major axis ($x$-axis), by the disk method: $V_1=\pi\displaystyle\int_{-a}^{a}y^2\,dx=\pi b^2\left[x-\dfrac{x^3}{3a^2}\right]_{-a}^{a}=\pi b^2\left(2a-\dfrac{2a}{3}\right)=\dfrac{4}{3}\pi ab^2$. …

The volume generated is $9\pi$ cubic units.

1. The triangle has vertices $(0,0)$, $(3,0)$, $(3,3)$; its sides are $y=0$ (base), $x=3$ (vertical side) and the hypotenuse joining $(0,0)$ to $(3,3)$, which is the line $y=x$.
2. For $0\le x\le3$, the region lies between $y=0$ and $y=x$.
3. Rotating about the $x$-axis, the volume is $V=\pi\int_0^3 y^2\,dx=\pi\int_0^3 x^2\,dx$.
4. $\int_0^3 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^3=\dfrac{27}{3}=9$. …