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 Post subject: Another Volumes of Revolution Question (SOLVED)  |  Posted: Fri Feb 24, 2012 7:39 pm

Original Member

Joined: Sat Aug 06, 2011 3:44 pm
Posts: 298
Location: UK

 A trickier one this time- the answer is obtained is slightly different from the one in the textbook.Here is the question: "Find the volume of the solid of revolution formed by rotating this region about the x-axis:$\displaystyle y = x\sin 3x \hspace{3mm};\: 0 \le x \le \frac{\pi}{3}$My working...$\displaystyle V = \pi \int_0^{\frac{\pi}{3}}x^2 \sin^2 3x\:dx \Rightarrow Let \: v = x^2 \hspace{2mm} \frac{du}{dx} = \sin^2 3x$$\displaystyle \sin^2 3x = \frac{1}{2} - \frac{1}{2}\cos 6x \Rightarrow \therefore u = \frac{1}{2}x - \frac{1}{12}\sin 6x\hspace{3mm} and\hspace{3mm} \frac{dv}{dx} = 2x$$ \displaystyle \Rightarrow \pi \int_0^{\frac{\pi}{3}} x^2 \sin^2 3x\: dx = \pi\left[x^2\left(\frac{1}{2}x - \frac{1}{12}\sin 6x\right)\right]^\frac{\pi}{3} _0$$\displaystyle - \pi\int_0^{\frac{\pi}{3}}2x\left(\frac{1}{2} x -\frac{1}{12} \sin 6x\right)\: dx$$ \displaystyle \Rightarrow \pi\int_0^{\frac{\pi}{3}} 2x\left(\frac{1}{2}x - \frac{1}{12}\sin 6x\right)\: dx$$\displaystyle \Rightarrow Let\: v = 2x \hspace{2mm} and \hspace{2mm} \frac{du}{dx} = \frac{1}{2} x - \frac{1}{12}\sin 6x$$ \displaystyle \Rightarrow \frac{dv}{dx} = 2 \hspace{2mm} u = \frac{1}{4}x^2 + \frac{1}{72}\cos 6x$$\displaystyle \Rightarrow \pi\int_0^{\frac{\pi}{3}} 2x\left(\frac{1}{2}x - \frac{1}{12}\sin 6x\right) \: dx$$ \displaystyle = \pi\left[2x\left(\frac{1}{4}x^2 +\frac{1}{72}\cos 6x\right)\right]^\frac{\pi}{3} _0$$\displaystyle - \pi\int_0^{\frac{\pi}{3}} \frac{1}{2} x^2 + \frac{1}{36}\cos 6x \: dx$$ \displaystyle \Rightarrow \pi\int_0^{\frac{\pi}{3}}x^2 \sin^2 3x = \pi\left[x^2\left(\frac{1}{2}x - \frac{1}{12}\sin 6x\right)\right]^{\frac{\pi}{3}} _ 0$$\displaystyle - \pi\left[2x\left(\frac{1}{4}x^2 + \frac{1}{72}\cos 6x\right)\right]^\frac{\pi}{3} _ 0$$ \displaystyle + \pi\int_0^{\frac{\pi}{3}} \frac{1}{2} x^2 + \frac{1}{36}\cos 6x\: dx$Then integrating that last term above and thus inputting the limits into each term gave me$\displaystyle \pi\left[\frac{\pi^3}{54} - \frac{2\pi\left(72\pi^2 + 36\right)}{7776} + \frac{\pi^3}{162}\right]$which simplifies to$\displaystyle \frac{\pi^2(48\pi^2 - 72)}{7776}$as the volume.Anyway, the answer given by the text book is$\displaystyle \frac{1}{324}\pi^2(2\pi^2 - 3)$which after going through my calculations again I realise is the same as the bloody textbook answer, now I feel like a right idiot for posting a whole thread which was actually corre t anyway *sigh*You can ignore this thread now if you like. _________________"Nature doesn't care what we call it, she just does it anyway".- Feynman
Prometheus
 Post subject: Re: Another Volumes of Revolution Question (SOLVED)  |  Posted: Sat Feb 25, 2012 5:46 pm

Original Member

Joined: Sun Aug 07, 2011 8:58 am
Posts: 309

 Haha... done that a few times myself.
iNow
 Post subject: Re: Another Volumes of Revolution Question (SOLVED)  |  Posted: Sat Feb 25, 2012 11:58 pm

Original Member

Joined: Thu Aug 04, 2011 11:40 pm
Posts: 5601
Location: Iowa

 It is kinda funny, but as Prometheus alludes... we've all done it. _________________iNow"[Time] is one of those concepts that is profoundly resistant to a simple definition." ~C. Sagan
x(x-y)
 Post subject: Re: Another Volumes of Revolution Question (SOLVED)  |  Posted: Sun Mar 04, 2012 8:07 pm

Original Member

Joined: Sat Aug 06, 2011 3:44 pm
Posts: 298
Location: UK

 Lol, I admit- it is funny, but quite annoying after typing a load of TeX which takes a while! _________________"Nature doesn't care what we call it, she just does it anyway".- Feynman
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