# Math Help - Surface area by revolution: Why is this comming out negative?

1. ## Surface area by revolution: Why is this comming out negative?

Find the area of the surface generated when the arc of the curve $x=\frac{3}{5}y^{3/5}-\frac{3}{4}y^{1/3}$ between y = 0 and y = 1 is revolved around (a) the y-axis, (b) the x-axis, and (c) the line y = -1.

I am stuck on the first one and thought that I should figure out what I am not getting before I press on. My answer keeps comming up negative, and I can't figure out why. Here is what I have:

$S=2\pi\int_{c}^{d}g(y)\sqrt{1+[g'(y)]^{2}}dy$
$S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{5/3}-\frac{3}{4}y^{1/3}\right)\sqrt{1+\left(y^{2/3}-\frac{1}{4}y^{-2/3}\right)^{2}}dy$
$S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{5/3}-\frac{3}{4}y^{1/3}\right)\left(y^{2/3}+\frac{1}{4}y^{-2/3}\right)dy$
$S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{7/3}-\frac{3}{5}y-\frac{3}{16}y^{-1/3}\right)dy$
$S=2\pi\left(\frac{9}{50}y^{10/3}-\frac{3}{10}y-\frac{9}{32}y^{2/3}\right)\mid_{0}^{1}$
$S=\frac{-321\pi}{400}$

Inverting the limits of integration (integrating from 1 to 0 instead of from 0 to 1) would make the answer positive, but is there a reason to do that? I so, could you explain why? Or is there some other mistake I am making? Like I said, I would like to get this sorted before I go on to the other problems. (I am not looking for help on the other parts -- b and c -- of the problem at this point, except for some general guidance, especially relative to what ever mistake I am making in this problem.)

2. See the graph below --note x = g(y) is negative so use -g(y) and go from there

3. Originally Posted by Calculus26
See the graph below --note x = g(y) is negative so use -g(y) and go from there
Ok, I see. Thanks so much for the help!