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Prometheus
 Post subject: And another one  |  Posted: Sun Feb 26, 2012 11:59 am

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

 Volumes of revolutions seem to be in vogue. I've just started learning about them and already getting confused by this simple looking one.Find the volume of the solid formed when the area enclosed by the curve $\displaystyle \frac {y}{x}=3-x^2$ and the x-axis performs one revolution about the x-axis.From this i get: $\displaystyle y=-x^2+3$ from which you can see the curve intersects the axis at $\displaystyle -\sqrt3$ and $\displaystyle \sqrt3$So, i make it:$\displaystyle \int_{-\sqrt3}^{\sqrt3}\:\pi(-x^2+3)^2. dx$Then plugging in the values:$\displaystyle \pi[-\frac {(\sqrt3)^3}{3}+3(\sqrt3)]^2-\pi[-\frac {(-\sqrt3)^3}{3}+3(-\sqrt3)]^2=0$Obviously wrong. Just hope i'm missing something basic rather than fundamental, so help appreciated.
Paper Wings
 Post subject: Re: And another one  |  Posted: Sun Feb 26, 2012 2:37 pm

Joined: Wed Aug 31, 2011 1:48 pm
Posts: 47

 The answer wouldn't happen to be, $\displaystyle \frac{48\pi}{5}\sqrt{3}$, would it?EDIT: No it's not, the lesson here being don't ignore the LHS of equations. Looks like x(x-y) has got it. Last edited by Paper Wings on Tue Feb 28, 2012 12:48 am, edited 1 time in total.
x(x-y)
 Post subject: Re: And another one  |  Posted: Mon Feb 27, 2012 9:01 pm

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

 $\displaystyle \frac{y}{x} = 3-x^2 \Rightarrow y = x(3-x^2)$$\displaystyle 0 = x(3-x^2) \Rightarrow x = -\sqrt{3}\:,\:x=0\:,\:x=\sqrt{3}So, \displaystyle V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} y^2 \: dx$$ \displaystyle y^2 = x^2(3-x^2)^2 \Rightarrow y^2 = 9x^2 -6x^4 + x^6$$\displaystyle \pi \int_{-\sqrt{3}}^{\sqrt{3}} 9x^2 -6x^4 + x^6\:dx = \pi\left[3x^3 - \frac{6}{5}x^5 + \frac{1}{7}x^7\right]^{\sqrt{3}}_{-\sqrt{3}}Subbing in the limits gives me \displaystyle \pi\left(9\sqrt{3} - \frac{54}{5}\sqrt{3} + \frac{27}{7}\sqrt{3}\right)$$ \displaystyle - \pi\left(-9\sqrt{3}+\frac{54}{5}\sqrt{3} - \frac{27}{7}\sqrt{3}\right)$Which yields$\displaystyle \frac{144\pi\sqrt{3}}{35}$as the solution to the volume of revolution.Hopefully that's right, maybe somebody else can confirm?Hope that helped! _________________"Nature doesn't care what we call it, she just does it anyway".- Feynman
Prometheus
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 1:07 am

Original Member

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

 Ah, i've been lying to you guys by mistyping the original equation, it was: $\displaystyle \frac{y}{x}=3-x$ which would make:$\displaystyle y=-x^2-3x \rightarrow x=0\:, x=3$So i've done it again, ironing out some mistakes, for hopefully the right answer (won't actually be given the answer til next week)$\displaystyle y^2= -x^3+3x^2$$\displaystyle \pi\int_0^3 \frac{-x^4}{4}+x^3dx$$ \displaystyle \pi(-\frac{3^4}{4}+3^3)-\pi(\frac{0^4}{4}+0^3) = \frac{198\pi}{4}$Since i made you guys do the cubic function i did it too - good to practice. I got the same as x(x-y) and a missed bedtime.
Paper Wings
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 7:20 am

Joined: Wed Aug 31, 2011 1:48 pm
Posts: 47

 Prometheus wrote:Ah, i've been lying to you guys by mistyping the original equation, it was: $\displaystyle \frac{y}{x}=3-x$ which would make:$\displaystyle y=-x^2-3x \rightarrow x=0\:, x=3$ don't you mean $\displaystyle y=3x - x^2$? making the roots -3 and 0.
Prometheus
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 11:16 am

Original Member

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

 Yes, that is what I meant $\displaystyle -x^2+3x$, how many typos am i going to make? Though i still make the roots to be 3 and 0. The homework is due in today, so i'll stick with what i've got. At least it's not zero.Thanks for the help, and i'll let you know the answer when i get it.
Paper Wings
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 12:54 pm

Joined: Wed Aug 31, 2011 1:48 pm
Posts: 47

 As you should...testing you i was...that's right...testing...
x(x-y)
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 2:46 pm

Original Member

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

 Well, in that case then- I get this$\displaystyle \frac{y}{x} = 3-x \Rightarrow y = x(3-x)$So, when y = 0 => x = 0 and x = 3$\displaystyle y^2 = x^2(3-x)^2 = 9x^2 -6x^3 + x^4$$\displaystyle V = \pi\int_{0}^{3} 9x^2 -6x^3 + x^4\: dx = \pi\left[3x^3 - \frac{3}{2}x^4 + \frac{1}{5}x^5\right]^{3}_{0}$$ \displaystyle \pi\left(3(27) - \frac{3}{2}(81) + \frac{1}{5}(243)\right)$Giving$\displaystyle \frac{81}{10}\pi$as the solution to the volume of revolution.I'm pretty sure that's right, we'll see... _________________"Nature doesn't care what we call it, she just does it anyway".- Feynman
Prometheus
 Post subject: Re: And another one  |  Posted: Tue Feb 28, 2012 9:36 pm

Original Member

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

 I'm glad to see those numbers. I changed my mind during break before hand-in and got the very same.
x(x-y)
 Post subject: Re: And another one  |  Posted: Sun Mar 04, 2012 1:22 pm

Original Member

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

 You could, of course, use integration by parts to solve this instead of multiplying the whole expression out. Here's how I'd do that:$\displaystyle \pi\int_{0}^{3} x^2(3-x)^2\:dx \Rightarrow Let\hspace{2mm}v = x^2\hspace{2mm} and \hspace{2mm} \frac{du}{dx}= (3-x)^2$$\displaystyle \frac{dv}{dx} = 2x \hspace{2mm} and \hspace{2mm} u = -\frac{1}{3}(3-x)^3$$ \displaystyle \Rightarrow \pi\int_{0}^{3} x^2(3-x)^2\:dx = \pi\left[-\frac{x^2}{3}(3-x)^3\right]^{3}_{0}$$\displaystyle + \pi\int_{0}^{3} \frac{1}{3}(3-x)^3 \times 2x\:dx$$ \displaystyle \pi\int_{0}^{3} \frac{1}{3}(3-x)^3 \times 2x\:dx \Rightarrow Let\hspace{2mm}v=2x\hspace{2mm}and\hspace{2mm}\frac{du}{dx}=\frac{1}{3}(3-x)^3$$\displaystyle \frac{dv}{dx}=2\hspace{2mm}and\hspace{2mm}u=-\frac{1}{12}(3-x)^4$$ \displaystyle \Rightarrow \pi\int_{0}^{3} \frac{1}{3}(3-x)^3 \times 2x\:dx = \pi\left[-\frac{x}{6}(3-x)^4\right]^{3}_{0}$$\displaystyle +\pi\int_{0}^{3}\frac{1}{6}(3-x)^4\:dx$$ \displaystyle \Rightarrow \pi\int_{0}^{3} x^2(3-x)^2\:dx = \pi\left[-\frac{x^2}{3}(3-x)^3\right]^{3}_{0} + \pi\left[-\frac{x}{6}(3-x)^4\right]^{3}_{0}$$\displaystyle +\pi\left[-\frac{1}{30}(3-x)^5\right]^{3}_{0}$And then inputting the limits yields$\displaystyle \frac{81}{10}\pi$as before. That's just another method you can use, it's probably more difficult than just multiplying out in this case- but for other integration questions it's very useful and much quicker. _________________"Nature doesn't care what we call it, she just does it anyway".- Feynman
Prometheus
 Post subject: Re: And another one  |  Posted: Fri Mar 23, 2012 2:09 pm

Original Member

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

 Yeah, that was the right answer. Cheers guys.
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