Volumes of Revolution

I was wondering if anybody could help me with this question on volumes of revolution:

"the region enclosed by both axes, the line x=2 and the curve y=1/8 x^2 + 2 is rotated about the y axis to form a solid, find the volume of this solid"

I was trying to work out a formula in terms of y for x squared and integrate this between 0 and 2.5 (the y value when x=2 is put into the formula) but apparently that's not how it is done, any help much appreciated :smile:

Reply 1

*bobo*

the volume is given by between 0and2 INT (pi)y^2 dx
so bet 0and 2 INT (pi)(0.125x^2 +2)^2 dx

Reply 2

Square

I think this is how you do it, it's worded abit weirdly....

Think of it graphically and in the positive quadrant only. the curve is going to be a U-shaped parabola and x=2 is going to be a horizontal line, where the two intersect is going to be your upper limits for your integration and where the turning point is will be your lower limit.

Then you use the formula:

$\displaystyle V=\pi\int_a^b x^2dy$

Reply 3

*bobo*

you can remember this by thinking of the solid formed as many cylinders of radius y and width dx. hence you want the sum of all the cylinders volumes which is INT pi y^2. dx

Reply 4

hamjam

I thought as it was rotated about the y axis and not the x axis I would not just be able to integrate y^2 and multiply it by pi as this would give me the volume of the solid formed if rotated about the x-axis? or am I just being slow today

Reply 5

Square

Cock, i got them mixed up, ignore me ! I'm the one being slow today.

Remember, you need to rearrange into the form: f(y)=x and the limits will be in terms of y as well, not x like they are normally.

Reply 6

*bobo*

ah, sorry.
in that case you integrate x^2 between y values achieved when you apply x=0 and x=2 to the equation

Reply 7

pyrolol

The moral of the story here is always to check the volume of revelution question with a fine toothed comb.

In my experience it seems to be the one where even excellent candidates make stupid mistakes, like using the wrong axis, missing out pi, of forgetting to square.

Reply 8

hamjam

*bobo*

ah, sorry.
in that case you integrate x^2 between y values achieved when you apply x=0 and x=2 to the equation

Thanks, this is what I thought I just realised I integrated between 0 and 2.5 and forgot to put the 0 into the formula,

y=1/8x^2+2
x^2 = 8(y-2)
put 0 and 2 in to get limits of 2 and 2.5
integrate to get [4x^2-16x] and use the limits to get the volume = 1*pi
could anyone check that for me?
as apparently it is 9pi :s-smilie:

(although this book has a habit of being wrong)

EDIT: just realised I have to take that away from the volume of the solid 2.5*2^2*pi which is 10pi -pi =9pi giving me the right answer, thanks for help :smile:

Reply 9

*bobo*