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x(x-y)
Post  Post subject: Volumes of Revolution Question  |  Posted: Wed Nov 23, 2011 10:54 pm
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Not to sound cocky, but I rarely get stuck on homework questions but this one (from the "OCR Advanced Mathematics Core 3 and 4" book, just in-case you're wondering) has got me scratching my head:

"The region enclosed by both axes, the line x = 2 and the curve

$ \displaystyle y=\frac{1}{8}x^{2} + 2$

is rotated about the y-axis to form a solid. Find the volume of this solid".

NOTE: The answer, according to the back of the book, is $ \displaystyle 9\pi$

So, this is what I've done so far... Sorry if I've made an obvious mistake...

$ \displaystyle x = 2 \Rightarrow y = \frac{1}{8}(2)^2 + 2 = \frac{5}{2}$

$ \displaystyle 8(y-2)^{\frac{1}{2}} = x$

$ \displaystyle V = \int_{0}^{\frac{5}{2}} \pi [8(y-2)]\: dy - \int_{0}^{\frac{5}{2}} \pi 2^2 \: dy$

$ \displaystyle \Rightarrow V = \int_{0}^{\frac{5}{2}} \pi (8(y-2) - 4) \: dy$

$ \displaystyle \Rightarrow \pi [4(y-2)^2 - 4y];\frac{5}{2}, 0$

... which does not give $ \displaystyle 9\pi$ as an answer for a Volume.

NOTE: I've attempted this many different ways, including subtracting the total volume of the "cuboidal area" from the answers obtained from integrating- but not to any avail!

Any help is appreciated, thanks!

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DrRocket
Post  Post subject: Re: Volumes of Revolution Question  |  Posted: Thu Nov 24, 2011 8:39 am
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x(x-y) wrote:
Not to sound cocky, but I rarely get stuck on homework questions but this one (from the "OCR Advanced Mathematics Core 3 and 4" book, just in-case you're wondering) has got me scratching my head:

"The region enclosed by both axes, the line x = 2 and the curve

$ \displaystyle y=\frac{1}{8}x^{2} + 2$

is rotated about the y-axis to form a solid. Find the volume of this solid".



You were on the right track but you got the sign and the integral wrong.

$ \displaystyle Vol = \pi \times 2^2 \times \frac {5}{2} - \int_2^{\frac{5}{2}} 8 \pi (y-2) dy = 9 \pi$

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x(x-y)
Post  Post subject: Re: Volumes of Revolution Question  |  Posted: Thu Nov 24, 2011 12:14 pm
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Ah yes, I see now- thanks for the help! I drew the graph quickly too- which helped to see what the solid of revolution would actually look like.

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