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November 12th, 2016, 06:15 PM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Surface Area of a Solid of Revolution
Given $\displaystyle y = x^3$ $\displaystyle SA = 2\pi\int y\cdot ds$ $\displaystyle ds = \sqrt{1+y'^2}=\sqrt{1+(3x^2)^2}=\sqrt{1+9x^4}$ The solid of revolution is rotated about the xaxis. Limits of integration are taken to be a = 0 and b = 1. $\displaystyle SA = 2\pi{\int^1_0} x^3 \sqrt{1+9x^4}\cdot dx$ Let $\displaystyle u = 1+9x^4$, $\displaystyle du = 36x^3\cdot dx$ New limits of integration: $\displaystyle b = \sqrt{1+9(1)^4} = \sqrt{10}$ $\displaystyle a = \sqrt{1+9(0)^4} = 1$ $\displaystyle SA = \frac{1}{18}\pi\int^{\sqrt{10}}_1 u\cdot du$ $\displaystyle SA = \frac{1}{18}\pi \left[\frac{u^2}{2}\right]^{\sqrt{10}}_1$ Substitute for u... Is this correct? Thank you in advance. Last edited by skipjack; November 13th, 2016 at 01:54 AM. 
November 12th, 2016, 06:25 PM  #2 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 
Looks like textbook example to me 
November 12th, 2016, 06:26 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
I don't see any problems with that. (Apart from the missing $\mathrm dx$ differentials in the working for $\mathrm ds$).
Last edited by greg1313; November 12th, 2016 at 06:50 PM. 
November 12th, 2016, 10:59 PM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\displaystyle SA \ = \ \frac{\pi}{36} (u^2)\bigg^{\sqrt{10}}_1$ Last edited by skipjack; November 13th, 2016 at 01:50 AM.  
November 13th, 2016, 04:32 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
"Could" rather than "should" in my opinion.


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area, revolution, solid, surface 
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