Le Volume Slicing (1 Viewer)

Aysce · Apr 12, 2012

Hello MX2ers, can you just verify what the dimensions of the annulus swept out of this diagram are?

I was thinking the outer radius is 2 and the inner radius is 2-x but when I did area and etc I got the wrong answer so yeahs.

http://www.mediafire.com/?gjoz94m8fd5wbua

SpiralFlex · Apr 12, 2012

$Take the full disc's radius as$ \;3-x.\; $The inner disc's radius has a constant radius of 2. We can use an annulus to find the volume.$

$\delta V= \pi (R^2-r^2) \delta y$

$V=\lim_{\delta y\rightarrow 0}\sum_{y=0}^{1}\pi*((3-x)^2-2^2)\delta y$

$=\lim_{\delta y\rightarrow 0}\sum_{y=0}^{1}\pi*(x^2-6x+5)\delta y$

$=\lim_{\delta y\rightarrow 0}\sum_{y=0}^{1}\pi*(y-6\sqrt{y}+5)\delta y$

$=\pi \int_{0}^{1} (y-6\sqrt{y}+5)\; dy$

$=\frac{3\pi}{2} \; \textup{units}^3$

Aysce · Apr 12, 2012

Err why would the radius be those two things?

EDIT: Realised. Note to self: need to draw diagrams better...

Thanks a lot.

SpiralFlex · Apr 12, 2012

When doing these questions always draw your diagrams as follows:

Edit: I'll do it after eating.

Aysce · Apr 12, 2012

SpiralFlex said:
When doing these questions always draw your diagrams as follows:

Edit: I'll do it after eating.

Nah I just drew it wrong.

And I want my cookies back!

Lolly_Polly · Apr 12, 2012

wow, so smart =">

Le Volume Slicing (1 Viewer)

Aysce

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SpiralFlex

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Aysce

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SpiralFlex

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Aysce

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Lolly_Polly

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