calculus

find the volume of solid inside the paraboloid z=9-x^2-y^2, outside the cylinder x^2+y^2=4 and above the xy-plane
1) solve using double integration of rectangular coordinate.
2) solve using double integration of polar coordinate
3)solve using triple intergation

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asked by AIRA
  1. First find out where z(x,y) intersects the x-y plane, which turns out to be a circle given by
    x²+y²=9
    So you would need evaluate the volume of the paraboloid above the x-y plane between the circles
    x²+y²=2² (cylinder, radius = 2)
    and
    x²+y²=3² end of volume above x-y plane, r=3.

    In polar coordinates, it would be
    ∫[0,2π]∫[2,3] z(x,y)rdrdθ
    z(x,y) can be converted to r,θ by the substitution
    x=rcosθ
    y=rsinθ
    so
    z(x,y)=9-x²-y²
    =9-r²(cos²θ+sin²θ)
    and the integral becomes
    Volume
    =∫[0,2π]∫[2,3] (9-r²)r dr dθ
    =∫[0,2π]∫[2,3]
    (9r-r³) dr dθ
    =2π[9r²/2-r^4/4] [2,3]
    =2π(25/4)
    =25π/2

    In rectangular coordinates, you could integrate over a semi-annulus from -3 to +3.

    Triple integration would be similar to (1) and (2), where z goes from 0 to z(x,y) or z(r,θ).

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  2. Why not this kind of answer?
    The cylinder intersects the paraboloid at z=5.

    The volume inside the paraboloid is

    v = ∫[0,2π] ∫[0,2] ∫[0,9-r^2] r dz dr dθ
    = ∫[0,2π] ∫[0,2] r(9-r^2) dr dθ
    = ∫[0,2π] 14 dθ
    = 28π

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    posted by AIRA
  3. Unless I am mistaken, the paraboloid looks like a mountain, tapering off to zero along a circle of radius 3.

    The cylinder has a radius of 2 and its axis is along the z-axis.

    So the volume of the paraboloid outside the cylinder is like you drilled a vertical hole of radius 2 through the mountain (of radius 3).

    I hope you can visualize the situation.

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  4. ohhh i see...
    so my answer just now is wrong??

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    posted by AIRA
  5. yes, sorry I only gave the volume of the paraboloid, and forgot to subtract the cylinder inside. Better go with MathMate's answer.

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    posted by Steve
  6. it's ok.. thank you..

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    posted by AIRA

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