Question

Someone please help me #Volume
Find volume of y=x^2, y=2-x^2, rotated through pi radians about the Y-axis.

Answer 1

I tried \[\pi\int\limits_{0}^{2} (\sqrt{2-y})^2-(\sqrt{y})^2 dy\] but I'm getting \[\pi(0)\] at the last, did I miss something?

Answer 2

why are you sqrting stuff?

Answer 3

rotating it by pi radians is the same as taking half of it and rotating it by 2pi radians ... a full circle due to its symmetry

Answer 4

cause its rotating by Y-axis? got it by changing these :y=x^2, y=2-x^2

Answer 5

a shell run would simply be:\[\int_{0}^{1}2\pi rh\]
such that r=x, and h is the difference between the function: (2-x^2)-(x^2)

Answer 6

you ran discs ... which is why you tried to invert them that fine too

Answer 7

so I can get the answer by putting in the correct values? but I'm still confuse why cant I get the answer by the way I did..

Answer 8

in your setup you subtracted the functions instead of splitting them up

Answer 9

you have a set of discs that runs from 0 to 1 for the bottom part

you have a set of discs that runs from 1 to 2 for the top part

top and bottom will produce equal volumes since they are symmetric, and you subtracted equal parts giving you a zero result :)

Answer 10

\[\pi\int\limits_{0}^{1} (\sqrt{2-y})^2~dy~+~\pi\int_{1}^{2}(\sqrt{y})^2 dy\]

Answer 11

ohh, I think I get what you meant, thank you so much! I'll look into it now :)

Answer 12

good luck ;)