Q.Find, by integration, the volume of the solid generated by revolving about the $x$-axis, the region enclosed by $y=e^{-2x}$, $y=0$, $x=0$ and $x=1$.

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: $\pi\displaystyle\int_0^a(ax-x^2)\,dx=\pi\left[\dfrac{ax^2}2-\dfrac{x^3}3\right]_0^a$.
4. Substitute $x=a$: $\pi\left(\dfrac{a\cdot a^2}2-\dfrac{a^3}3\right)=\pi\left(\dfrac{a^3}2-\dfrac{a^3}3\right)$.
5. Common denominator: $\dfrac{a^3}2-\dfrac{a^3}3=\dfrac{3a^3-2a^3}6=\dfrac{a^3}6$.
6. So $V=\dfrac{\pi a^3}6$.

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 $3$.
3. By the cone volume formula, $V = \dfrac13\pi r^2h = \dfrac13\pi(3)^2(3) = \dfrac13\pi\cdot27 = 9\pi$.
4. Verify by the disc method of integration: $V = \pi\displaystyle\int_0^3 y^2\,dx = \pi\int_0^3 x^2\,dx = \pi\left[\dfrac{x^3}{3}\right]_0^3 = \pi\cdot\dfrac{27}{3} = 9\pi$.

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: $V_1=\pi\displaystyle\int_{-a}^{a}y^2\,dx=\pi b^2\left[x-\dfrac{x^3}{3a^2}\right]_{-a}^{a}=\pi b^2\left(2a-\dfrac{2a}{3}\right)=\dfrac{4}{3}\pi ab^2$.
3. By the same method (swap roles of $x,y$), the volume revolved about the minor axis ($y$-axis) is $V_2=\dfrac{4}{3}\pi a^2 b$.
4. Ratio: $V_1:V_2=ab^2:a^2b=b:a$.

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. $\int_0^3 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^3=\dfrac{27}{3}=9$.
5. So $V=9\pi$.
6. Options (a) $18\pi$, (c) $36\pi$ are off by a factor of $2$ or $4$; (b) $2\pi$ is far too small for this region.