# 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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