The Student Room Group

Volumes of revolution

In question 7, for the rotation I chose Pi/2 rather than pi. But the books answer does use pi. Why? image.jpg
Original post by maggiehodgson
In question 7, for the rotation I chose Pi/2 rather than pi. But the books answer does use pi. Why? image.jpg


Choose pi/2 for what ?
Reply 2
Original post by RDKGames
Choose pi/2 for what ?


When you integrate x squared it’s usual to use pi to multiply the definite integral but as the area is symmetrical it wouldn’t have to rotate through 2pi to make a solid so I thought to use pi.
Reply 3
I think you need to tell everyone a bit more of what you did. For example, did you integrate in the y direction or the x direction. It is an E/P questions and so is expected to have a trick to it.
Original post by maggiehodgson
When you integrate x squared it’s usual to use pi to multiply the definite integral but as the area is symmetrical it wouldn’t have to rotate through 2pi to make a solid so I thought to use pi.


The formula to rotate a curve x=f(y)x=f(y) about the y axis a full turn of 2π2\pi radians to generate a volume VV is:

V=πabx2 dyV = \pi \displaystyle \int_a^b x^2 \ dy

The symmetry of the graph in y-axis does not matter, you take the x>0 part and rotate it all the way around the y-axis.
(edited 10 months ago)
Reply 5
Original post by nerak99
I think you need to tell everyone a bit more of what you did. For example, did you integrate in the y direction or the x direction. It is an E/P questions and so is expected to have a trick to it.


The question does say the rotation is about the y axis. I integrated x squared but used a multiplier of pi / w because of symmetry. However I’ve been told now to always use pi.
Reply 6
Original post by RDKGames
The formula to rotate a curve x=f(y)x=f(y) about the y axis a full turn of 2π2\pi radians to generate a volume VV is:

V=πabx2 dyV = \pi \displaystyle \int_a^b x^2 \ dy

The symmetry of the graph in y-axis does not matter, you take the x>0 part and rotate it all the way around the y-axis.


Well, there you go. I was thinking that just a half rotation would form a solid. Thanks
Reply 7
Original post by maggiehodgson
Well, there you go. I was thinking that just a half rotation would form a solid. Thanks


If you imagine that region R being a flat piece of card, then rotating through pi/2 would mean it would be sticking out of the page with its edge facing you - it wouldn't form a complete solid of revolution.
Reply 8
Original post by davros
If you imagine that region R being a flat piece of card, then rotating through pi/2 would mean it would be sticking out of the page with its edge facing you - it wouldn't form a complete solid of revolution.


But if you look at question 8 then that would need to rotate a full circle to make a solid but in question 7 the shape turns on the y axis and the right hand side swivels to the left and the left to the right. I am imagining the shape rotates clockwise. I must have a misunderstanding of what it all means.
Reply 9
Original post by maggiehodgson
But if you look at question 8 then that would need to rotate a full circle to make a solid but in question 7 the shape turns on the y axis and the right hand side swivels to the left and the left to the right. I am imagining the shape rotates clockwise. I must have a misunderstanding of what it all means.

The picture could be a bit misleading for a volume of revolution question, though for the curve
y^3 +x^2 - 2y = 4
its correct to draw it as pictured as for each vaue of y you have two solutions
x = +/- sqrt(4 + 2y - y^3)

In this case, we need to know the radius to get the volume of a thin disc, so the radius rotated though 2pi gives the area and multiplied by the height dy gives
volume of thin disc = pi r^2 dy
and
r^2 = x^2 = 4 + 2y - y^3
It doesnt matter which of +/- solutions it corresponds to. If youre unsure, maybe explicitly state what the radius (squared) is when you set up the integral.
(edited 10 months ago)
Original post by mqb2766
The picture could be a bit misleading for a volume of revolution question, though for the curve
y^3 +x^2 - 2y = 4
its correct to draw it as pictured as for each vaue of y you have two solutions
x = +/- sqrt(4 + 2y - y^3)

In this case, we need to know the radius to get the volume of a thin disc, so the radius rotated though 2pi gives the area and multiplied by the height dy gives
volume of thin disc = pi r^2 dy
and
r^2 = x^2 = 4 + 2y - y^3
It doesnt matter which of +/- solutions it corresponds to. If youre unsure, maybe explicitly state what the radius (squared) is when you set up the integral.

Thanks. My assumption that the region shaded would turn around the y axis is incorrect.
Reply 11
Original post by maggiehodgson
Thanks. My assumption that the region shaded would turn around the y axis is incorrect.


Its correct that R is rotating about the y axis as stated in the question, but whats incorrect is how you then map it to the volume of revolution formula. The formula is based on a radius rotating 2 pi to get the area/volume of a thin disc, and in this case, the radius represents half of R, as in the other figures/questions, so the distance from the centre (y axis) to the curve.

As a slight aside (inspred by notnek), when you evaluate the integral, it simply evaluates to
4*2*pi
so it has the same volume as a cylinder of radius 2 (radius^2=4) and height 2. This matches the given figure. With a bit of foresight you might notice that the
2y - y^3
cancel, or has zero area, over [0,2], so youre just left with the constant 4.
(edited 10 months ago)
Original post by mqb2766
Its correct that R is rotating about the y axis as stated in the question, but whats incorrect is how you then map it to the volume of revolution formula. The formula is based on a radius rotating 2 pi to get the area/volume of a thin disc, and in this case, the radius represents half of R, as in the other figures/questions, so the distance from the centre (y axis) to the curve.

As a slight aside (inspred by notnek), when you evaluate the integral, it simply evaluates to
4*2*pi
so it has the same volume as a cylinder of radius 2 (radius^2=4) and height 2. This matches the given figure. With a bit of foresight you might notice that the
2y - y^3
cancel, or has zero area, over [0,2], so youre just left with the constant 4.


If I had integrated between -2 and 2 rather than 0 and 2 then I would have been able to use pi/2?
Reply 13
Original post by maggiehodgson
If I had integrated between -2 and 2 rather than 0 and 2 then I would have been able to use pi/2?


No in general ... but yes iin this case as notnek noticed yesterday and hence the comment above ...

The [0,2] and [-2,2] refer to integrating along the y-axis, so the height of the shape. In general, the shape of the curve below the x-axis is irrelevant (not in R) and so in general its wrong. However, in this case it works as the integral over [-2,0] equals the integral over [0,2] so doubling (by integrating over [-2,2]) and then halving (dividing pi by 2) works. It only works in this case as there is no y^2 term so the point of inflection occurs at y=0 and the average value of r^2 = 4 in both intervals, for the reason in the previous post.
(edited 10 months ago)
Thank you all.

I'm happy now that I understand all comments and advice
Reply 15
Original post by maggiehodgson
Thank you all.

I'm happy now that I understand all comments and advice

As youve not uploaded your working, its may be irrelevant, but Id "always" sketch the curve(s) and the thin disc youre considering to make expllicit how you translate the question to the formula. So something like this guy does
https://www.youtube.com/watch?v=lBSLPUbYFsU&ab_channel=MichelvanBiezen
Original post by mqb2766
As youve not uploaded your working, its may be irrelevant, but Id "always" sketch the curve(s) and the thin disc youre considering to make expllicit how you translate the question to the formula. So something like this guy does
https://www.youtube.com/watch?v=lBSLPUbYFsU&ab_channel=MichelvanBiezen

Thanks for the video link. In all the questions in the book, so far, each has a diagram. I am the sort of person who needs a visual so it is something I would do if I weren't given one. Diagrams do help don't they.
Reply 17
Original post by maggiehodgson
Thanks for the video link. In all the questions in the book, so far, each has a diagram. I am the sort of person who needs a visual so it is something I would do if I weren't given one. Diagrams do help don't they.


Agreed, and even though they have given the curve, Id sketch it myself and mark on the thin disc, the radius and the integration limits etc, so everything is clear. For instance, it would be easy to read off the integration limits as being the [-2,2] on the x-axis? But this would be the wrong variable and also the function actually bulges beyond those limits.
Reply 18
Original post by maggiehodgson
The question does say the rotation is about the y axis. I integrated x squared but used a multiplier of pi / w because of symmetry. However I’ve been told now to always use pi.

I don't know if this is in your spec (although I'm pretty sure I've seen it needed in a TSR thread marked A-level), but as a general principle there are two ways of finding the volume of revolution - one where you integrate along the y-axis and one where you integrate along the x-axis.

As you've done, you can imagine that for each value of y there's a very thin (width dy) circular disc of radius x, giving you πx2dy\pi \int x^2\, dy.

But you can also imagine that for each value of x there's a very thin (width dx) cylindical shell of radius x and height y, giving you 2πxydx2\pi \int xy \, dx.

In the first case you need x as a function of y, in the second, y as a function of x. Occasionally it's *much* easier to write y as a function of x in which case you really want to use the 2nd approach.

[Conversely, for this particular problem, writing y as a function of x is hideous, so you definitely want the first approach].

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