Washer method rotating around horizontal line (not x-axis), part 2 | AP Calculus AB | Khan Academy

[@proanimator.]

Sal is awesome

[@yo0yo0yo0]

to rotate a function over another function you can simply use the function as your axis and then intigrate by that axis Meaning youll have to use the delta of the axis funciton as your dx. to calculate the du simply the derivative of y=mx+c. du=m dx.

[@senguo9131]

you may use factorization at the last phase; dont know if it may simplify but may circumvent that pile of calculations

[@albumalbum68]

Thank you for such clear explanation, but probobably there is a misspilling here: x^2-2x = x(x-2) , not x(x-3), so zeros will be 0 and 2

[@GodsNut]

What you said about "x^2-2x= x(x-2)" is true, but you forgot that he set "x= x^2-2x", to set them equal to each other (or as he said it, when is "x" equal to "x^2-2x", then he subtracted an "x" from both sides, leaving it as 0= x^2-3x, then he factored out the "x." Hope that helped.

[@zain1612]

surely its pi*(integral of (y^2)) with respect to to x

[@MrHaipingyang]

I guess you may need to use polar coordinates.

[@MrHaipingyang]

Using DOTS might be easier?