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Need help with integration question?

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.
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.
Reply 2
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?
Reply 3
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?
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 π27(x3)2.dx\displaystyle \pi \int_2^7 (x^3)^2 .dx
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.
(edited 7 years ago)
Reply 6
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
Reply 7
Original post by RDKGames
Well the process of integration depends on what you're integrating.

Since it's about the x-axis you have π27(x3)2.dx\displaystyle \pi \int_2^7 (x^3)^2 .dx


Is this the correct answer?
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
Reply 8
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!
(edited 7 years ago)
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 x[0,2]x \in [0,2] so this post is technically still incorrect.
to integrate you raise the power then divide by the new power ( and add c if there are no limits )
Reply 11
Original post by RDKGames
Well the process of integration depends on what you're integrating.

Since it's about the x-axis you have π27(x3)2.dx\displaystyle \pi \int_2^7 (x^3)^2 .dx


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
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)!
Reply 13
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
Reply 14
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.
(edited 7 years ago)
Reply 15
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 :wink:

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