volume about the y axis. (1 Viewer)

[atBondi]

Ok, so area bounded by curves y = x^2 and y = x +2 is rotated about the x-axis.

I have tried it so many time and i cant do it lol, so frustrating can anyone help?
but is the general approach for this find the volume of one of the parts and then subtract another part?

 

[life92]

First of all you need to find the points of intersection

x^2 = x + 2

x^2 - x - 2 = 0
(x-2)(x+1)=0
x = -1, 2

Now to find the volume about the x axis, its V = pi S y^2 dx
But here, since we're dealing with 2 graphs, we need to do the upper graph minus the lower graph. If we draw a diagram then its y=x+2 on top.

Therefore,
V = pi S (x+2)^2 - (x^2)^2 dx
= pi S (x+2)^2 - x^4 dx
= pi [(x+2)^3/3 - x^5 / 5] [-1=>2]
= pi [ (64/3 - 32/5) - (1/3 +1/5) ]
= 74 pi / 5

I think that's right.
I sometimes make mistakes when typing this out o_o
Btw, why is the title, volume about the y axis if its about the x axis o_o

 

[b00m]

Ok, so area bounded by curves y = x^2 and y = x +2 is rotated about the x-axis.

I have tried it so many time and i cant do it lol, so frustrating can anyone help?
but is the general approach for this find the volume of one of the parts and then subtract another part?

I think your problems can be attributed to your general confusion of the question. Is it around the y-axis? (as suggested by the thread title) or is it along the x-axis.. as stated in your post? lol.

edit: fek, was pointed out 3 days before me..

i got roughly 14.4, which is the same answer as life92's

 

[thongetsu]

yeah is is rotated about the x or y axis?