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

    Since it's about the x-axis you have \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
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    (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!
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    (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 \in [0,2] so this post is technically still incorrect.
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    to integrate you raise the power then divide by the new power ( and add c if there are no limits )
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    (Original post by RDKGames)
    Well the process of integration depends on what you're integrating.

    Since it's about the x-axis you have \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
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    (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)!
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    (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
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    (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.
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    (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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