Thread: [SOLVED] Surface Area - check my work?

1. [SOLVED] Surface Area - check my work?

Find the area of the surface obtained by rotating the following curve about the x-axis:

$y = x^3$, $0 \leq x \leq 3$

Here is my work, which I'm pretty sure is correct, but wrong in Webassign:

$2\pi \int\limits^{3}_{0} (x^3) \sqrt{1+9x^4} dx$

$\frac{\pi}{18} \int\limits^{729}_{1} \sqrt{u} du$

$= \frac{\pi}{18} \biggl[\frac{2}{3} u^{\frac{3}{2}} \biggr]^{729}_{1}$

$= \frac{\pi}{27} \biggl[729^{\frac{3}{2}} - 1 \biggr]$

Can someone please check my work and let me know where the error is?
Thanks!! Molly

2. Originally Posted by mollymcf2009
Find the area of the surface obtained by rotating the following curve about the x-axis:

$y = x^3$, $0 \leq x \leq 3$

Here is my work, which I'm pretty sure is correct, but wrong in Webassign:

$2\pi \int\limits^{3}_{0} (x^3) \sqrt{1+9x^4} dx$

$\frac{\pi}{18} \int\limits^{7{\color{red}30}}_{1} \sqrt{u} du$

$= \frac{\pi}{18} \biggl[\frac{2}{3} u^{\frac{3}{2}} \biggr]^{7{\color{red}30}}_{1}$

$= \frac{\pi}{27} \biggl[7{\color{red}30}^{\frac{3}{2}} - 1 \biggr]$

Can someone please check my work and let me know where the error is?
Thanks!! Molly
Note my corrections in red.