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 June 20th, 2014, 07:51 AM #1 Newbie   Joined: Jun 2014 From: paris Posts: 5 Thanks: 0 The volume of the solid of revolution rotated about the x-axis Hi, could someone please explain how to tackle the question below. Find the volume of the solid of revolution obtained when the graph of f(x)= sec x + tan x, from x=−[pie]/3 to x = [pie] is rotated about the x-axis. Give your answer to four decimal places. Thank you for taking the time!
 June 20th, 2014, 08:14 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus pap.gif $\displaystyle V=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\pi f(x)^2\;\mathbb{d}x=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\pi \left(\sec(x)+\tan(x)\right)^2\;\mathbb{d}x\approx 15.1859$
 June 20th, 2014, 08:21 AM #3 Newbie   Joined: Jun 2014 From: paris Posts: 5 Thanks: 0 Thank you ZardoZ, Would you mind explaining how you integrated (sec(x) + tan(x))^2 dx , as this is where I got stuck in the first place. Thank you again for elaborating and looking forward to your reply! E
June 20th, 2014, 08:25 AM   #4
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 Originally Posted by Einstein Thank you ZardoZ, Would you mind explaining how you integrated (sec(x) + tan(x))^2 dx , as this is where I got stuck in the first place. Thank you again for elaborating and looking forward to your reply! E
(sec⁡x+tan⁡x)^2=sec^2⁡x+2secxtanx+tan^2⁡(x )=∫sec^2⁡(x)dx+2∫sec(x) tan(x)dx+∫tan^2⁡(x) dx

 June 20th, 2014, 08:45 AM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus $\begin{eqnarray}\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\sec(x)+\tan(x )\right)^2\;\mathbb{d}x&=&\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)^2+2\sec(x) \tan(x)+\tan(x)^2\;\mathbb{d}x\\&=&\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)^2+2\sec(x) \tan(x)+\sec(x)^2\boxed{-1}\;\mathbb{d}x\\&=&2\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)^2\;\mathbb{d }x+2\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x) \tan(x)\;\mathbb{d}x-\frac{2\pi}{3}\\&=&2\tan(x)\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}+2\cdot 0 -\frac{2\pi}{3}\\&=&4\sqrt{3}-\frac{2\pi}{3}\end{eqnarray}$ You owe me for this $\rm \LaTeX$ writting....... Thanks from Einstein Last edited by ZardoZ; June 20th, 2014 at 09:35 AM.
 June 20th, 2014, 09:28 AM #6 Newbie   Joined: Jun 2014 From: paris Posts: 5 Thanks: 0 Thank you ZardoZ, really helpful explanation However, I'm still unclear how you calculated - 2pie/3, I think I'm missing sth. Could you explain that part? Thank you again! E
 June 20th, 2014, 09:32 AM #7 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus It was that -1 that dissapeared in the next step $-\displaystyle \int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}1\;\mathbb{d}x=-x\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=-\frac{\pi}{3}+\left(-\frac{\pi}{3}\right)=-\frac{2\pi}{3}$. Look again in the previous post, I put it in a box. Thanks from Einstein Last edited by ZardoZ; June 20th, 2014 at 09:36 AM.
 June 20th, 2014, 10:43 AM #8 Newbie   Joined: Jun 2014 From: paris Posts: 5 Thanks: 0 ZerdoZ, great!!! I hope you don't mind one last question about 2 * 0, just before the finishing line. I calculated -4(3)^½ rather than 0 ..... It would be great if you could go through that part, please. Much appreciated Thank you again for everything, really helpful! E
 June 20th, 2014, 10:56 AM #9 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Twinkle twinkle little star, why didn't you tell me from the beginning that you cannot integrate? There are two ways. 1. $\displaystyle \int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)\tan(x)\; \mathbb{d}x=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\sec(x)\right) '\;\mathbb{d}x=\sec(x)\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=0$ 2. If $f(x)=\sec(x)\tan(x)$, then in $\left[-\frac{\pi}{3},\frac{\pi}{3}\right]$ f is strictly increasing and f(0)=0 with f(-x)=-f(x) (odd). $\displaystyle \int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)\tan(x)\; \mathbb{d}x=\int\limits_{0}^{\frac{\pi}{3}}\sec(x) \tan(x)\; \mathbb{d}x-\int\limits_{0}^{\frac{\pi}{3}}\sec(x)\tan(x)\; \mathbb{d}x=0$ Thanks from Einstein Last edited by ZardoZ; June 20th, 2014 at 10:58 AM.
 June 20th, 2014, 11:05 AM #10 Newbie   Joined: Jun 2014 From: paris Posts: 5 Thanks: 0 Great help! Have a great weekend, Zardoz E

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### find volume of solid of revolution by rotating thru 2pi around x axis secx tanx

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