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# Need help with integration question? watch

1. Volume of revolutions:
Find the volume generated when the region bounded by y=x^3 and the x-axis between x=2 and x=7 is rotated through 360° about the x-axis.
2. (Original post by Sniper21)
Volume of revolutions:
Find the volume generated when the region bounded by y=x^3 and the x-axis between x=2 and x=7 is rotated through 360° about the x-axis.
Use the formula for rotation about the x-axis.
3. (Original post by RDKGames)
Use the formula for rotation about the x-axis.
Do I need to integrate it first? Reduce the power by 1 and then divide by the new
power?
4. (Original post by tajtsracc)

Since it's rotated about the x-axis, use the top equation. Make sure you square both sides of y = x3 to make it y2.
Please show me the step by step method?
5. (Original post by Sniper21)
Do I need to integrate it first? Reduce the power by 1 and then divide by the new
power?
Well the process of integration depends on what you're integrating.

Since it's about the x-axis you have
6. (Original post by Sniper21)
Please show me the step by step method?
I'll start off for you.

Square both sides of y = x3. This gives you y2 = (x3)2 = x6.

Sub this into the first volume equation.
∫πy2 dx = ∫π(x6) dx

Take the pi to the outside for now.
π∫(x6) dx.

Now integrate x6. After that use your limits 7 and 2.
7. (Original post by tajtsracc)
I'll start off for you.

Square both sides of y = x3. This gives you y2 = (x3)2 = x6.

Sub this into the first volume equation.
∫πy2 dx = ∫π(x6) dx

Take the pi to the outside for now.
π∫(x6) dx.

Now integrate x6. After that use your integrals 7 and 2.
Can you double check if my answer is right?
y=x^3

v= π y^2 dx
So v=
2
∫π (x^3)^2 dx = π x^6 dx (upper limit is 2, lower limit is 0)
0

= π (x^7/7) (upper limit 2, lower limit 0)
= π (2^7/7) - (0^7/7) =128π/7 units^3
8. (Original post by RDKGames)
Well the process of integration depends on what you're integrating.

Since it's about the x-axis you have
y=x^3

v= π y^2 dx
So v=
2
∫π (x^3)^2 dx = π x^6 dx (upper limit is 2, lower limit is 0)
0

= π (x^7/7) (upper limit 2, lower limit 0)
= π (2^7/7) - (0^7/7) =128π/7 units^3
9. (Original post by tajtsracc)
Your working out seems fine, though I'm not sure why your limits have changed? Weren't they originally 7 and 2?
Correction:
y=x^3

v= π y^2 dx
So v=
7
∫π (x^3)^2 dx = π x^6 dx (upper limit is 7, lower limit is 2)
2

= π (x^7/7) (upper limit 7, lower limit 2)
= π (2^7/7) - (0^7/7) =128π/7 units^3

I know how to do it now. Thank you, I really appreciate your help!
10. (Original post by Sniper21)
Correction:
y=x^3

v= π y^2 dx
So v=
7
∫π (x^3)^2 dx = π x^6 dx (upper limit is 7, lower limit is 2)
2

= π (x^7/7) (upper limit 7, lower limit 2)
= π (2^7/7) - (0^7/7) =128π/7 units^3

I know how to do it now. Thank you, I really appreciate your help!
This is not a correction - all you did was change the limits while keeping the working out with domain so this post is technically still incorrect.
11. to integrate you raise the power then divide by the new power ( and add c if there are no limits )
12. (Original post by RDKGames)
Well the process of integration depends on what you're integrating.

Since it's about the x-axis you have
Correction:
y=x^3

v= π y^2 dx
So v=
2
∫π (x^3)^2 dx = π x^6 dx (upper limit is 7, lower limit is 2)
0

= π (x^7/7) (upper limit 7, lower limit 2)
= π (2^7/7) - (0^7/7) =128π/7 units^3
13. (Original post by Sniper21)
Correction:
y=x^3

v= π y^2 dx
So v=
2
∫π (x^3)^2 dx = π x^6 dx (upper limit is 7, lower limit is 2)
0

= π (x^7/7) (upper limit 7, lower limit 2)
= π (2^7/7) - (0^7/7) =128π/7 units^3
Your last line is wrong (you are still using the limits 2 and 0)!
14. (Original post by DFranklin)
Your last line is wrong (you are still using the limits 2 and 0)!
Is this right?
∫ πy² dx from x = 2 to x = 7
= ∫ π(x³)² dx from x = 2 to x = 7
= ∫ πx⁶ dx from x = 2 to x = 7
= 1/7 πx⁷
= 1/7π( 7⁷ - 2⁷)
= 823415π/7
≈ 369548 cubic metres
15. (Original post by Sniper21)
Is this right?
∫ πy² dx from x = 2 to x = 7
= ∫ π(x³)² dx from x = 2 to x = 7
= ∫ πx⁶ dx from x = 2 to x = 7
= 1/7 πx⁷
= 1/7π( 7⁷ - 2⁷)
= 823415π/7
≈ 369548 cubic metres
That's correct. One small thing: the units should only be cubic metres if the question refers to metres.
16. (Original post by notnek)
That's correct. One small thing: the units should only be cubic metres if the question refers to metres.
It came up on the exam today, and I got it right! I put the correct units in as well. Thanks for your reply

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