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June 20th, 2014, 08:51 AM   #1
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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!
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June 20th, 2014, 09:14 AM   #2
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$\displaystyle V=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\pi f(x)^2\;\mathbb{d}x=\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\pi \left(\sec(x)+\tan(x)\right)^2\;\mathbb{d}x\approx 15.1859$
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June 20th, 2014, 09:21 AM   #3
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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
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June 20th, 2014, 09:25 AM   #4
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Quote:
Originally Posted by Einstein View Post
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
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June 20th, 2014, 09:45 AM   #5
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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.......
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Last edited by ZardoZ; June 20th, 2014 at 10:35 AM.
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June 20th, 2014, 10:28 AM   #6
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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
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June 20th, 2014, 10:32 AM   #7
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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.
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Last edited by ZardoZ; June 20th, 2014 at 10:36 AM.
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June 20th, 2014, 11:43 AM   #8
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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
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June 20th, 2014, 11:56 AM   #9
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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).

$\displaystyle \int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec(x)\tan(x)\; \mathbb{d}x=\int\limits_{0}^{\frac{\pi}{3}}\sec(x) \tan(x)\; \mathbb{d}x-\int\limits_{0}^{\frac{\pi}{3}}\sec(x)\tan(x)\; \mathbb{d}x=0$
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Last edited by ZardoZ; June 20th, 2014 at 11:58 AM.
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June 20th, 2014, 12:05 PM   #10
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Great help! Have a great weekend, Zardoz

E
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