Q.The surface area of the solid of revolution of the region bounded by $x^2 + y^2 = 4$, $x = -2$ and $x = 2$ about $x$-axis is :

Revolving the semicircle $x^2+y^2=4$ (for $-2\le x\le2$) about the $x$-axis gives a sphere of radius $2$ with surface area $16\pi$.

1. The curve $x^2+y^2=4$ is a circle of radius $2$ centred at the origin, and $x=-2,x=2$ are its extreme $x$-values (the endpoints of a diameter).
2. Revolving this full circle (or equivalently its upper semicircle over the whole range $-2\le x\le2$) about the $x$-axis produces a complete sphere of radius $r=2$. …

Using the surface-of-revolution formula on $y=2x$ over $[0,2]$ gives surface area $8\sqrt5\,\pi$.

1. The curve is $y=2x$ for $x\in[0,2]$, revolved about the $x$-axis.
2. The surface area formula is $S=2\pi\displaystyle\int_a^b y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$.
3. Here $\dfrac{dy}{dx}=2$, so $\sqrt{1+2^2}=\sqrt5$, a constant. …

Surface area of revolution of $y=2x$ about the $x$-axis from $x=0$ to $x=2$ is $8\sqrt5\pi$.

1. The surface area generated by revolving $y=f(x)$ about the $x$-axis from $x=a$ to $x=b$ is $S=2\pi\int_a^b y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$
2. Here $y=2x$, so $\dfrac{dy}{dx}=2$ and $1+\left(\dfrac{dy}{dx}\right)^2=1+4=5$.
3. So $S=2\pi\int_0^2 2x\sqrt5\,dx=2\pi\sqrt5\int_0^2 2x\,dx$.
4. $\int_0^2 2x\,dx=[x^2]_0^2=4$.
5. Hence $S=2\pi\sqrt5(4)=8\sqrt5\pi$. …