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 Calculus Calculus Math Forum

 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 x-axis. 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 Thanks from hyperbola November 12th, 2016, 06:26 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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 $\mathrm ds$). Thanks from hyperbola Last edited by greg1313; November 12th, 2016 at 06:50 PM. November 12th, 2016, 10:59 PM   #4
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Quote:
 Originally Posted by hyperbola $\displaystyle SA = \frac{1}{18}\pi \left[\frac{u^2}{2}\right]^{\sqrt{10}}_1$ Substitute for u... Is this correct? Thank you in advance.
Before substituting into u, you should simplify it first:

$\displaystyle SA \ = \ \frac{\pi}{36} (u^2)\bigg|^{\sqrt{10}}_1$

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: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra "Could" rather than "should" in my opinion. Tags area, revolution, solid, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post xl5899 Calculus 3 December 16th, 2015 10:13 AM rob747uk Calculus 4 April 18th, 2015 05:56 AM mhuan66 Calculus 0 April 18th, 2015 02:01 AM ehh Calculus 3 February 10th, 2013 02:50 PM Jet1045 Calculus 19 March 28th, 2012 01:49 PM

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