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|dw:1349809459606:dw|

just this part ?

i think i need to consider the negative x values too

|dw:1349809625381:dw|

|dw:1349809619079:dw|

this part rotated around the y-axis

so i think we should separate into positive x and negative x

agree so far?

yes

wait .. i think that we dont need the positive x part at all

since the revolution of the negative part will cover it

i was thinking in use \[\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx\]

them after i calculate this two parts i sum them?

yeah, after the rotation it will cover the positive part

ok a got it! thank you!

so if we add the positive part we get an extra for the answer and its bad..

yeah... that question is fun rsrsr

(a-b)^2 = (b-a)^2

because the function 3-x² is in far of y = x in relation to the y-axis?

(a-b)^2 = (b-a)^2 !!

ohh ok i got it!

thank you!!

;)

oh wait!!

there is something wrong there!

sorry for that.. so long havent done such things

because we like subtracting the inner disc from the larger disc