Question

revolve the region in first quadrant bounded by y = x^2 , y = x , about line x = 2, using washer method

Answer 1

\[\pi \int\limits_{0}^{4}(2-x)^{2}-(2-x ^{2})^{2}\]

Answer 2

thats wrong

Answer 3

noooooooooo

Answer 4

srry its been a long time.

Answer 5

the region is from (0,0) to (1,1)

Answer 6

oh sorry sorry forgot about the bounds. assumed that x22 was also a bound

Answer 7

also youre using dy , not x

Answer 8

ok nvm i rambl.ed

Answer 9

now you will die

Answer 10

ure supposed to use dx?

Answer 11

dy

Answer 12

i was using DY

Answer 13

since its horizontal integration

Answer 14

just the wrong values

Answer 15

no its wrong in other ways!!!

Answer 16

ok lets do it again: \[\pi \int\limits_{0}^{1} (2-x)^{2}-(2-x ^{2})^{2}\]

Answer 17

this HASSSSS TO BE RIGHT

Answer 18

eh eh?

Answer 19

no

Answer 20

its pi* integral ( y - 2)^2 - (sqrt y - 2)^2