calculus

Find the volume of the solid when the region enclosed by the curves y=((25-(x^2))^(1/2))and y=3 is revolved about the x-axis.

I already have the answer = (256*pi)/3, but don't know how to set up.

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  1. From x=-4 to x=4, the y=[(25-(x^2))^(1/2)] curve is above y > 3.
    From x = -5 to -4 and 4 to 5, y <3

    It is not clear what the x limits of integration are, nor how the two different regions, outside and inside y=3, are to be treated.

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  2. I set both y equal to each other to find the points of integration.

    squared both sides
    (((25-(x^2))^(1/2))=3)^2
    then becomes
    (x^2)=25-9
    Result
    x= +(14^(1/2) and -(14^(1/2)

    I don't know what to use, as in the disk or washer method.

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  3. First find the volume generated when the circle
    x^2 + y^2 = 25 from -4 to 4 is rotated (from symmetry I will go from 0 to 4 and double)
    V = 2π[integral] (25 - x^2)dx from 0 to 4
    = 2π [25x - (1/3)x^3] from 0 to 4
    = 2π(100 - 64/3) = 472π/3

    then we will "hollow out" a cylinder with radius 3 and height of 8 (from -4 to 4)
    that volume is π(9)(8) = 72π

    The volume of our rotated solid is
    472π/3 - 72π
    = 472π/3 - 216π/3
    = 256π/3 as required

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  4. Thanks

    Helped a lot

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  5. Congrats to Reiny for that; I was hoping he would take it. I hope you see my point that there are two additional volumes between the two curves from x = -5 to -4 and from 4 to 5. Those volumes are cylindrical on the OUTside. Apparently they expected you to ignore them

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