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shaon0

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Consider the region bounded by y=3-x^2, y=x+x^2 and x=-1. The point P iis the intersection of y=3-x^2 and y=x+x^2 in the 1st quadrant.
i) Sketch the region. DONE
ii) Find the point P. DONE
iii) Hence, find the volume of the solid generated when this regiom is rotated about y=3.

I don't know how to approach part iii). Any help will be appreciated. Thanks
 

GUSSSSSSSSSSSSS

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use slices perpendicular to x axis
cross sections will be annulus with outer radius equal to 3 - (x + x^2)
and inner radius equal to 3 - (3 - x^2)
integrate between 1 and -1


btw the answer i got was 1 is that right ???
 
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MC Squidge

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i havent done volumes yet, i cant work it out. so im gonna presume dd24 is right
 

gurmies

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Lol @ volume = 0. I haven't done this topic, but I assume that's wrong. Tsk tsk - failure to run a dimensional analysis.
 

GUSSSSSSSSSSSSS

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dammit my latex thingy isnt working =[[[

Lol @ volume = 0. I haven't done this topic, but I assume that's wrong. Tsk tsk - failure to run a dimensional analysis.
you assume correct sire




and i still stand by my first answer of 1 as i just did it a second time
..........but then again i might be doin something fundamentally wrong xD
 

GUSSSSSSSSSSSSS

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gah fuck too tired ur answer was rite study freak

for me soooooooooooo many misreadings of my writing =[[[[[[[[[[[[[[[[
 

study-freak

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gah fuck too tired ur answer was rite study freak

for me soooooooooooo many misreadings of my writing =[[[[[[[[[[[[[[[[
well, this question required too much simple algebra (simple, but easy to make mistakes)
I made those mistakes more than once in my 4u half yearlies lol...
 

kwabon

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could someone please show or tell what exactly is the region you are rotating about y=3, because i am still struggling to figure what exactly is the region.

sorry for the inconvenience
 

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