Author Topic: Calling out Brightsky (Read 4156 times) Tweet Share

enwiabe · « **Reply #15 on:** July 27, 2012, 03:05:14 am »

Quote from: Special At Specialist on July 18, 2012, 05:54:24 pm

Well that's awkward then... the teacher is going to hate me for this

No teacher should hate you for inquiring! It's still a good question, you just needed to realise that it's not quite on the course

Mao · « **Reply #16 on:** July 27, 2012, 03:31:06 am »

Quote from: Special At Specialist on July 18, 2012, 04:00:18 pm

You asked a question which has really got me stumped. Not only me, but my entire specialist class, including the teacher! We spent a whole period discussing this question as a class and couldn't manage to solve it. Now the teacher is really eager to see the solution and so am I.
Please show me how this is done without the use of a calculator:

Quote from: brightsky on June 26, 2012, 07:41:21 pm
have a go at this:

find the volume of the solid obtained when y = e^x between y = 1 and y = 4 is rotated around the y-axis.

Alternatively, we can do this with integration by shell.

$V = \sum 2\pi r\times height\times thickness = \int_{x_0}^{x_1} 2\pi x f(x)\, dx$

In this case, we need to transform the volume so the base is sitting on the x axis. This is easily achieved with a translation and a reflection. The new function is thus $y=4-e^x$ . This corresponds to x=0 to x= $\log(4)$ .

The integral is

$V = \int_{0}^{\log(4)} 2\pi x (4-e^x )\, dx$

...which still requires integration by parts. Nevermind.

enwiabe · « **Reply #17 on:** July 27, 2012, 06:37:17 am »

Quote from: Mao on July 27, 2012, 03:31:06 am

...which still requires integration by parts. Nevermind.

Ugh Spec Maths is so annoying for not having integ by parts. So many interesting integrals arise from that technique, and it's really not so difficult a concept! It's on the syllabus in the NSW 'equivalent' (quote marks because it's really much harder than spec)

paulsterio · « **Reply #18 on:** July 27, 2012, 07:46:33 pm »

Hmm, NSW mathematics is much more real maths, but much less applications, just from seeing their exams.

Jenny_2108 · « **Reply #19 on:** July 27, 2012, 11:23:22 pm »

Quote from: Mao on July 27, 2012, 03:31:06 am

$V = \sum 2\pi r\times height\times thickness = \int_{x_0}^{x_1} 2\pi x f(x)\, dx$

Oh man, this looks interesting! But why $V = \sum 2\pi r\times height\times thickness$
I thought $V = 2\pi r\times height\times thickness$

Mao · « **Reply #20 on:** July 28, 2012, 03:38:29 am »

Quote from: Jenny_2108 on July 27, 2012, 11:23:22 pm

Quote from: Mao on July 27, 2012, 03:31:06 am
$V = \sum 2\pi r\times height\times thickness = \int_{x_0}^{x_1} 2\pi x f(x)\, dx$

Oh man, this looks interesting! But why $V = \sum 2\pi r\times height\times thickness$
I thought $V = 2\pi r\times height\times thickness$

Each individual cylindrical shell has the volume $dV=2\pi r\times height\times thickness$ , the total volume of all the individual cylindrical shells is thus $V=\sum dV$

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Author Topic: Calling out Brightsky (Read 4156 times) Tweet Share

enwiabe

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