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bleakarcher · Sep 10, 2011

how the fuk do u do q5) b)

http://4unitmaths.com/2003WRpaper.pdf

thoth1 · Sep 10, 2011

you find the volume of the whole sphere and call that V1. then you find volume of cylinder and call it V2.
therefore Vtot of remaining sphere = V1-V2

posting solution in a sec

K4M1N3 · Sep 10, 2011

thoth1 said:
you find the volume of the whole sphere and call that V1. then you find volume of cylinder and call it V2.
therefore Vtot of remaining sphere = V1-V2

posting solution in a sec

But your not accounting for the part of the cylinder that extrudes the sphere...

thoth1 · Sep 10, 2011

crap my way doesnt work.

bleakarcher · Sep 10, 2011

i tryed rotating the semi-circle y=sqrt[2-x^2] about the line y=-1.

bleakarcher · Sep 10, 2011

lol. i did this yesterday at night as i was half a sleep. i figured lol. u must rotate the semi-circle y=sqrt[4-x^2] about the line y=-1.

bleakarcher · Sep 10, 2011

there should be a pi there in the exact volume, so i do not think so. but u just rotate the region bounded by the curve y=sqrt[4-x^2] and the x-axis about the line y=-1 to obtain the volume.

thoth1 · Sep 10, 2011

$x^{2}+y^{2}=9$

$x^{2}=9-y^{2}$

$at x =1 y = 2\sqrt{2}$

$\delta V_{1} = \pi \int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2}dy = 2\pi \int_{0}^{2\sqrt{2}}9-y^{2}dy$

$\delta V_{2}=\pi (1)(4\sqrt{2})=4\sqrt{2}\pi$

$\delta V_{TOT}=\delta V _{1}-\delta V_{2}$

$\delta V_{TOT}=\frac{64\sqrt{2}\pi}{3}$

Drongoski · Sep 10, 2011

Good work thoth1

I just worked out the same answer: $\frac {64 \sqrt 2 \pi}{3}$ cubic units.

In uni this would be done more directly via a double integral.

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