Posted by **nicko** on Monday, January 3, 2011 at 6:16am.

find the volume formed by revolving the parabola y=4-x^2 in the first quadrant:

a.) about x=0

b.) about x=2

- calculus -
**Reiny**, Monday, January 3, 2011 at 9:33am
a)

after making the diagram, take horizontal discs, so

volume = π[integral] x^2 dy from 0 to 4

= π[integral](4-y)dy from 0 to 4

= π [4y - y^2/2] from 0 to 4

= π( 16 - 8 - 0) = 8π

b) since you only want to rotate the part in the first quadrant, look at your diagram.

I would find the volume of the cylinder with radius 2 and height of 4, the "hollow out" the part contained by y = 4-x^2, x=2 and y=4

we need the radius or the x of that part, which is 2 - √(4-y)

but we will need x^2 which is

4 - 2√(4-y) + 4-y

= 8 - y - 2(4-y)^(1/2)

this will have to be integrated to give

8y - y^2/2 + (4/3)(4-y)^(3/2)

see if you can finish it from here.

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