anonymous
  • anonymous
revolve the region in first quadrant bounded by y = x^2 , y = x , about line x = 2, using washer method
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\pi \int\limits_{0}^{4}(2-x)^{2}-(2-x ^{2})^{2}\]
anonymous
  • anonymous
thats wrong
anonymous
  • anonymous
noooooooooo

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anonymous
  • anonymous
srry its been a long time.
anonymous
  • anonymous
the region is from (0,0) to (1,1)
anonymous
  • anonymous
oh sorry sorry forgot about the bounds. assumed that x22 was also a bound
anonymous
  • anonymous
also youre using dy , not x
anonymous
  • anonymous
ok nvm i rambl.ed
anonymous
  • anonymous
now you will die
anonymous
  • anonymous
ure supposed to use dx?
anonymous
  • anonymous
dy
anonymous
  • anonymous
i was using DY
anonymous
  • anonymous
since its horizontal integration
anonymous
  • anonymous
just the wrong values
anonymous
  • anonymous
no its wrong in other ways!!!
anonymous
  • anonymous
ok lets do it again: \[\pi \int\limits_{0}^{1} (2-x)^{2}-(2-x ^{2})^{2}\]
anonymous
  • anonymous
this HASSSSS TO BE RIGHT
anonymous
  • anonymous
eh eh?
anonymous
  • anonymous
no
anonymous
  • anonymous
its pi* integral ( y - 2)^2 - (sqrt y - 2)^2

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