# Vol. of Revolution

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#1
How would you do this for a loop, eg.

The area between the curves:

y = 5
y = x^2 + 2

Is rotated about the x-axis. Find the volume of the loop produced.

Because for another question, I found the area of the shape, squared it and multiplied it by pi, but this obviously can't be done on every occasion - the further from the x-axis, the larger the volume will be. So maybe when you square the area and times by pi you find the vol. of the shape when you rotate that shape about the x-axis when it is toughing the x-axis.

But how would you do it it it wasn't touching the x-axis? And is this needed for P2?
0
16 years ago
#2
(Original post by mik1a)
How would you do this for a loop, eg.

The area between the curves:

y = 5
y = x^2 + 2

Is rotated about the x-axis. Find the volume of the loop produced.

Because for another question, I found the area of the shape, squared it and multiplied it by pi, but this obviously can't be done on every occasion - the further from the x-axis, the larger the volume will be. So maybe when you square the area and times by pi you find the vol. of the shape when you rotate that shape about the x-axis when it is toughing the x-axis.

But how would you do it it it wasn't touching the x-axis? And is this needed for P2?
work out the volume for the curve y=5 to which you shoudl subtract the volume generated under the curve of y=x^2+2
0
16 years ago
#3
(Original post by mik1a)
How would you do this for a loop, eg.

The area between the curves:

y = 5
y = x^2 + 2

Is rotated about the x-axis. Find the volume of the loop produced.

Because for another question, I found the area of the shape, squared it and multiplied it by pi, but this obviously can't be done on every occasion - the further from the x-axis, the larger the volume will be. So maybe when you square the area and times by pi you find the vol. of the shape when you rotate that shape about the x-axis when it is toughing the x-axis.

But how would you do it it it wasn't touching the x-axis? And is this needed for P2?
I'll be damned if this is right
0
#4
(Original post by hihihihi)
work out the volume for the curve y=5 to which you shoudl subtract the volume generated under the curve of y=x^2+2
Ahh, that makes perfect sense. Thankyou.

I'll try it now:

y = 5
y = x^2 + 2

5 = x^2 + 2
x^2 = 3
x = +/- sqrt3
So length is 2sqrt2

Volume by y=5:

pi*5*5*2*sqrt2 = 50.pi.sqrt2 (vol. of cylinder)
~ 272.07

Volume by x^2 + 2

= (sqrt3/-sqrt3){ pi.(x^2 + 2)^2 dx
= 2(sqrt3/0){ pi.(x^2 + 2)^2 dx
= 2(sqrt3/0){ pi.(x^4 + 4x^2 + 4) dx
= 2pi[x^5/5 + 4x^3/3 + 4x](sqrt3/0)
= 2pi[(sqrt3)^5/5 + 4(sqrt3)^3/3 + 4(sqrt3)]
= 2pi[3^(5/2)/5 + (4/3)(3^3/2) + 4*3^(1/2)]
~ 106.65

Difference:

272.06 - 106.65 = 165.42 cube units (2d.p.)
Hmm...
0
16 years ago
#5
I got 82.71

The cylinder has radius 5, height root3, therefore volume = 136.0349523...

then subtract the volume of the curve rotated about the x axis, I got the same volume as kimoni, so the volume of the "loop" would be 82.71. But I might have missed sumthing there...have you got the answer??
0
#6
(Original post by Katie Heskins)
I got 82.71

The cylinder has radius 5, height root3, therefore volume = 136.0349523...

then subtract the volume of the curve rotated about the x axis, I got the same volume as kimoni, so the volume of the "loop" would be 82.71. But I might have missed sumthing there...have you got the answer??
I made it up. But I'm sure the height is 2 root 3, because the points where they intersect are at +/- root 3, so the height is -rt3 - +rt3?
0
16 years ago
#7
(Original post by mik1a)
How would you do this for a loop, eg.

The area between the curves:

y = 5
y = x^2 + 2

Is rotated about the x-axis. Find the volume of the loop produced.

Because for another question, I found the area of the shape, squared it and multiplied it by pi, but this obviously can't be done on every occasion - the further from the x-axis, the larger the volume will be. So maybe when you square the area and times by pi you find the vol. of the shape when you rotate that shape about the x-axis when it is toughing the x-axis.

But how would you do it it it wasn't touching the x-axis? And is this needed for P2?
omg those are some ugly numbers

i actually got the same as u mikia, approx. 165
0
16 years ago
#8
(Original post by mik1a)
I made it up. But I'm sure the height is 2 root 3, because the points where they intersect are at +/- root 3, so the height is -rt3 - +rt3?
haha yes you are quite right, I just assumed that it was between root 3 and 0! But u wanted it between -root 3 and root 3 hence why your answer is twice mine 0
16 years ago
#9
(Original post by Katie Heskins)
haha yes you are quite right, I just assumed that it was between root 3 and 0! But u wanted it between -root 3 and root 3 hence why your answer is twice mine yay 0
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