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# Volumes of rotation watch

1. Find the volume of revolutions when the area enclosed by the curves 𝑦 = 2𝑥^2
and 𝑦 = 𝑥^3 is rotated through 360° about the 𝑦-axis.

I got 256pi/35 but the markscheme says 16pi/5

2. (Original post by stolenuniverse)
Find the volume of revolutions when the area enclosed by the curves 𝑦 = 2𝑥^2
and 𝑦 = 𝑥^3 is rotated through 360° about the 𝑦-axis.

I got 256pi/35 but the markscheme says 16pi/5

Post your working and we'll tell you where you went wrong.
3. (Original post by stolenuniverse)
Find the volume of revolutions when the area enclosed by the curves 𝑦 = 2𝑥^2
and 𝑦 = 𝑥^3 is rotated through 360° about the 𝑦-axis.

I got 256pi/35 but the markscheme says 16pi/5

I also got the same answer as you -- Please let me know if you find out where you went wrong
4. (Original post by Jasondazza)
I also got the same answer as you -- Please let me know if you find out where you went wrong
I'm starting to get the feeling the mark-scheme is wrong.. but i will tell you if it's not

EDIT: it's the y-axis *facepalm* XD
5. (Original post by RDKGames)
Post your working and we'll tell you where you went wrong.
sorry if the photo is bad
Attached Images

6. (Original post by stolenuniverse)
sorry if the photo is bad
The question states it is rotated about the y-axis therefore you need to use the formula . You've used the one for rotation about the x-axis.
7. You're finding the volume when you rotate around the x-axis, you need the volume when you rotate about the y-axis.
8. (Original post by RDKGames)
The question states it is rotated about the y-axis therefore you need to use the formula . You've used the one for rotation about the x-axis.
Thank you!!!
9. (Original post by RDKGames)
The question states it is rotated about the y-axis therefore you need to use the formula . You've used the one for rotation about the x-axis.
You can also use (where here y should be the difference between the 2 curves) which avoids needing to rewrite the original expressions to get x as a function of y.

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