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Prometheus


Original Member
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}=3x^2$ and the xaxis performs one revolution about the xaxis.
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


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(xy) has got it.
Last edited by Paper Wings on Tue Feb 28, 2012 12:48 am, edited 1 time in total.





x(xy)


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

$ \displaystyle \frac{y}{x} = 3x^2 \Rightarrow y = x(3x^2)$
$ \displaystyle 0 = x(3x^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(3x^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


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}=3x$ which would make:$ \displaystyle y=x^23x \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(xy) and a missed bedtime.





Paper Wings


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}=3x$ which would make:$ \displaystyle y=x^23x \rightarrow x=0\:, x=3$ don't you mean $ \displaystyle y=3x  x^2$? making the roots 3 and 0.





Prometheus


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


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

As you should...testing you i was...that's right...testing...





x(xy)


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} = 3x \Rightarrow y = x(3x)$
So, when y = 0 => x = 0 and x = 3
$ \displaystyle y^2 = x^2(3x)^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


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 handin and got the very same.





x(xy)


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(3x)^2\:dx \Rightarrow Let\hspace{2mm}v = x^2\hspace{2mm} and \hspace{2mm} \frac{du}{dx}= (3x)^2$
$ \displaystyle \frac{dv}{dx} = 2x \hspace{2mm} and \hspace{2mm} u = \frac{1}{3}(3x)^3$
$ \displaystyle \Rightarrow \pi\int_{0}^{3} x^2(3x)^2\:dx = \pi\left[\frac{x^2}{3}(3x)^3\right]^{3}_{0}$$ \displaystyle + \pi\int_{0}^{3} \frac{1}{3}(3x)^3 \times 2x\:dx$
$ \displaystyle \pi\int_{0}^{3} \frac{1}{3}(3x)^3 \times 2x\:dx \Rightarrow Let\hspace{2mm}v=2x\hspace{2mm}and\hspace{2mm}\frac{du}{dx}=\frac{1}{3}(3x)^3$
$ \displaystyle \frac{dv}{dx}=2\hspace{2mm}and\hspace{2mm}u=\frac{1}{12}(3x)^4$
$ \displaystyle \Rightarrow \pi\int_{0}^{3} \frac{1}{3}(3x)^3 \times 2x\:dx = \pi\left[\frac{x}{6}(3x)^4\right]^{3}_{0}$$ \displaystyle +\pi\int_{0}^{3}\frac{1}{6}(3x)^4\:dx$
$ \displaystyle \Rightarrow \pi\int_{0}^{3} x^2(3x)^2\:dx = \pi\left[\frac{x^2}{3}(3x)^3\right]^{3}_{0} + \pi\left[\frac{x}{6}(3x)^4\right]^{3}_{0}$$ \displaystyle +\pi\left[\frac{1}{30}(3x)^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


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

Yeah, that was the right answer. Cheers guys.





