# [SOLVED] Surface Area - check my work?

• Mar 10th 2009, 07:25 PM
mollymcf2009
[SOLVED] Surface Area - check my work?
Find the area of the surface obtained by rotating the following curve about the x-axis:

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

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

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

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

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

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

Can someone please check my work and let me know where the error is?
Thanks!! Molly
• Mar 10th 2009, 07:32 PM
Chris L T521
Quote:

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

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

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

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

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

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

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

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

Note my corrections in red.