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P3 Integration watch

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    1) The finite region bounded by the curve with equation y=tan0.5x, the line x=pi/2 and the coordinate axes is rotated through 2pi radians about the x-axis. Show that the volume generated is [pi(4-pi)]/2.

    2) By rotating the semi-circle for which y> or equal to 0 from the circle with equation x^2 + y^2 = a^2 completely about the x-axis, show that the volume of a sphere, of radius a, is [4pi(a^3)]/3.

    3) A triangular region is bounded by the line with equation y = rx/h, where r and h are positive constants, the line x=r and the x-axis. By considering the complete rotation of this triangle about the x-axis, show that the volume of a cone of height h and base radius r is [pi(r^2)h]/3.

    4) An ellipse is given by x=3cost, y=2sint, 0<or= t <or= 2pi.

    a) Find the area of the finite region bounded by the ellipse and the positive x-and y-axes.

    b) This region is rotated completely about the y-axis to form a solid of revolution. Find the volume of this solid.

    Answers:

    4a) 3pi/2.
    4b) 12pi.

    Cheers
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    1.
    V = pi ∫ tan²(x/2) dx
    = pi ∫ sec²(x/2) - 1 dx
    = pi [2tan(x/2) - x]^{\pi / 2}_{0}
    = pi [(2 - pi/2) - (0 - 0)]
    = pi (4 - pi)/2

    2.
    Find x-incercepts by setting y=0:
    x = ±a
    So:
    V = pi ∫ a² - x² dx
    = pi [a²x - (x³)/3]^{a}_{-a}
    = pi [(a³ - (a³)/3) - (-a³ + (a³)/3)]
    = pi (4/3) a³

    3.
    V = pi (r/h)² ∫ x² dx
    = pi (r²/3h²) [ x³ ]^{h}_{0}, since the height of the cone we want is h.
    = pi (r²/3h²) . h³
    = (pi r² h)/3

    4. Are you sure you typed it out correctly? I can't seem to get those answers.
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    (Original post by dvs)
    1.
    V = pi ∫ tan²(x/2) dx
    = pi ∫ sec²(x/2) - 1 dx
    = pi [2tan(x/2) - x]^{\pi / 2}_{0}
    = pi [(2 - pi/2) - (0 - 0)]
    = pi (4 - pi)/2

    2.
    Find x-incercepts by setting y=0:
    x = ±a
    So:
    V = pi ∫ a² - x² dx
    = pi [a²x - (x³)/3]^{a}_{-a}
    = pi [(a³ - (a³)/3) - (-a³ + (a³)/3)]
    = pi (4/3) a³

    3.
    V = pi (r/h)² ∫ x² dx
    = pi (r²/3h²) [ x³ ]^{h}_{0}, since the height of the cone we want is h.
    = pi (r²/3h²) . h³
    = (pi r² h)/3

    4. Are you sure you typed it out correctly? I can't seem to get those answers.
    Cheers for the 3 answers

    I've checked it again, and I did type it out correctly :confused:
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    One question - why didn't you square the whole thing at the start of each of the 3 questions? If you want to find the volume, shouldn't you square all the contents before you integrate it?
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    (Original post by Aristotle)
    ...

    4) An ellipse is given by x=3cost, y=2sint, 0<or= t <or= 2pi.

    a) Find the area of the finite region bounded by the ellipse and the positive x-and y-axes.

    b) This region is rotated completely about the y-axis to form a solid of revolution. Find the volume of this solid.

    Answers:

    4a) 3pi/2.
    4b) 12pi.

    Cheers
    4/
    a)
    A = int y dx
    = int 2sint.(-3sint) dt {pi/2, 0}
    = -6 int sin²t dt {pi/2, 0}
    = -3 int (1 - cos2t) dt {pi/2, 0}
    = -3[t - (1/2)sin2t]{pi/2, 0}
    = -3{(0 - 0) - (pi/2 - 0)}
    = 3pi/2
    =====

    b)
    for this part I get
    V = 8pi
    =====

    Anyone else ?
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    (Original post by Aristotle)
    One question - why didn't you square the whole thing at the start of each of the 3 questions? If you want to find the volume, shouldn't you square all the contents before you integrate it?
    dvs did "square" everything.
    He used the formula

    V = pi int y² dx

    each time.
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    (Original post by Fermat)
    b)
    for this part I get
    V = 8pi
    =====

    Anyone else ?
    I get that too! A mistake in the book perhaps?
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    (Original post by dvs)
    I get that too! A mistake in the book perhaps?
    Obviously

    4/
    b)

    eqn of ellipse is

    (x/3)² + (y/2)² = 1
    y²/4 = 1 - x²/9
    y² = 4(1 - x²/9)

    V = pi int y² dx
    V = pi int 4(1 - x²/9) dx
    V = 4pi int (1 - x²/9) dx
    V = 4pi[x - x^3/27] { 0,3}
    V = 4pi{(3 - 3^3/27) - (0 0 0)}
    V = 4pi{3 - 1}
    V = 8pi
    =====
 
 
 
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