find the volume of the solid obtained by rotating about the y-axis the region bounded by the curve y=2x^2-x^3 and the x-axis. How do I do this?
find the the x and y intercepts
ok.
y = x^2(2-3x) right?
y=0 at 0 and 3/2 right?
no y=x^2(2-x)
thats right lol.... i transposed that 3 dint i :)
so we have bounds of x=0 and x=2 then right?
yes.
we need to get these in terms of y to calculate the y axis spin; lets do an inverse
ok sounds good.
x = 2y^2 -y^3 ; solve for y :)
wolfram might help with this :)
good idea. I will get back to you a little later on this I have a baseball game.
are you sure its spun around the y axis? and not the x axis?
let me read up on the shell method; its prolly easier with this setup..
2pi [S] x[f(x)] dx ; from [0,2]
x(2x^2-x^3) = 2x^3 -x^4 soo... we integrate that to get; (5x^4 -2x^5)/10. At 0 it is zero; so we just worry about the 2
i get 8/5 as the volume of the torus....if i did it right lol
so i am going to try this on firefox..... and it seems to be working good so far :)
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