Find the area of the surface generated when the given curve is revolved about the x-axis
The part of the curve $y=12x-2$ between the points $(\frac{5}{12},3)$ and $(\frac{13}{12},11)$
I understand how to solve this when I am dealing one point, but I am confused then there are two points, I hope someone can guide me through this question.
I used the formula:
$$S=\int_{a}^{b} 2π \left(f(x)\right)\sqrt{1+f'(x)^2} $$
$$S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{1+(12)^2} $$
$$S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{145} $$
$$[2π( \left(12x-2\right)\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6x^2-2x\right)\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6\frac{5}{12}^2-2\frac{5}{12}\right)\sqrt{145})-(6\frac{13}{12}^2-2\frac{13}{12}\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6\frac{24}{144}-\frac{10}{12}\right)\sqrt{145})-((6\frac{169}{144}-\frac{26}{12})\sqrt{145})] $$
$$[2π( \left(-\frac{48}{3}\right)\sqrt{145})-((-\frac{858}{144})\sqrt{145})] $$
Best Answer
Since $\displaystyle S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{145} $, letting $u=12x-2$ and $du=12\;dx$ gives
$\displaystyle\frac{2\pi\sqrt{145}}{12}\int_3^{11} u\;du=\frac{\pi\sqrt{145}}{6}\left[\frac{u^2}{2}\right]_3^{11}=\frac{\pi\sqrt{145}}{12}(121-9)=\frac{28\pi\sqrt{145}}{3}$.
Alternatively, using $x=\frac{1}{12}(y+2),\;\; 3\le y\le 11$ gives
$\displaystyle S=\int_3^{11}2\pi y\sqrt{1+\frac{1}{144}}dy=\frac{2\pi\sqrt{145}}{12}\int_3^{11}y\;dy$.