Find the area of the surface obtained by revolving the given curve about the x-axis. on

Question

Find the area of the surface obtained by revolving the given curve about the x-axis.  on

Quizplus · Accepted Answer

The given curve is the graph of $y=x^2+2$ from $x=-1$ to $x=1.$ When this curve is revolved around the $x$-axis, we obtain a solid with circular cross sections. The radius of each circular cross section is equal to the distance from the $x$-axis to the curve, i.e. $r(x)=x^2+2.$ Therefore, the area of the surface of the solid is given by the formula:

$$2\pi \int_{-1}^{1} r(x) \sqrt{1+\left(\frac{dr}{dx}ight)^2} \,dx$$

Since $r(x)=x^2+2,$ we have $\frac{dr}{dx}=2x,$ so $\left(\frac{dr}{dx}ight)^2=4x^2.$ Therefore,

\begin{align*}
2\pi \int_{-1}^{1} r(x) \sqrt{1+\left(\frac{dr}{dx}ight)^2} \,dx &= 2\pi \int_{-1}^{1} (x^2+2)\sqrt{1+4x^2} \,dx \
&= \pi \int_{-2}^{2} u\sqrt{1+u^2} \,du \qquad (	ext{where } u=2x)\
&= \pi \cdot \frac{2}{3}(1+u^2)^{3/2} \bigg|_{-2}^{2} \
&= \boxed{\pi\left(8\sqrt{5}-\frac{16\sqrt{5}}{3}ight)}.
\end{align*}

Find the Area of the Surface Obtained by Revolving the Given