Calculus

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We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please? =) Thank you!

1) Find the volume of the solid formed when the region bounded by curves y=x^3 + 1, x= 1, and y=0 is rotated about the x-axis.

My answer is 23/14.

2) Find the volume of the solid of revolution obtained by revolving the region bounded by y=1/x and the lines x=pi/8 and x=pi/2 around the x-axis.

I got 6/pi for this one.

3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.

My answer: 5/12

Can you tell me if I did them wrong, and if I did do them wrong what I Did wrong?

  • Calculus -

    standard way to form volume if spun around x-axis

    y = x^3 + 1 ---> this becomes the radius of your disk

    we need the x-intercept,
    let x^3 + 1 = 0
    x = -1 ---> this becomes our left boundary

    Volume = π∫(x^3 + 1)^2 dx from x = -1 to x = 1
    = π∫(x^6 + 2x^3 + 1) dx
    = π [ x^7/7 + x^4/2 + x] from -1 to 1
    = π( 1/7 + 1/2 + 1 - ( -1/7) + 1/2 - 1)
    = P π( 2/7 + 2)= 16π/7

    for #2,
    y = 1/x
    so y^2 = 1/x^2

    V = π∫1/x^2 dx from π/8 to π/2
    = π [ - 1/x ] from π/8 to π/2
    = π ( -2/π - (-8/π)
    = -2+8 = 6

    #3, first you need the intersection:

    x^3 + x^2 = 2x^2 + 2x
    x^3 - x^2 - 2x = 0
    x(x^2 - x - 2) = 0
    x(x-2)(x+1) = 0

    they intersect at x = -1, x=0 and x=2
    So you must find the volume separately from
    x = -1 to x = 0 , and then from x = 0 to x = 2

    from -1 to 0 , the cubic is the upper curve, while from
    0 to 2, the parabola is the upper curve
    So you have
    V = π∫( (x^3 + x^2)^2 - (2x^2 + 2x)^2 ) dx from -1 to 0 + π∫( (2x^2 + 2x)^2 - (x^3 + x^2)^2 ) dx from 0 to 2

    = etc.

  • Calculus -

    Thank you very much Reiny. I understand it a bit better now.

  • Calculus -

    I did it and I get 4653pi/105 but this is wrong. Have I calculated incorrectly?

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