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## Homework Statement

evaluate volume of paraboloid z = x

^{2}+ y

^{2}between the planes z=0 and z=1

## The Attempt at a Solution

i figured we would need to rearrange so that F(x,y,z) = x

^{2}+ y

^{2}- z

then do a triple integral dxdydz of the function F. the limits for the first integral dz would be z=1 and z=0. and i dont know what the other limits would be (y1,y2 and x1, x2?)

but this first integral gave an answer of -1/2. this would mean that the volume would end up being zero which i dont think is right.

then i thought that maybe i should say the function is z(x,y) = x

^{2}+ y

^{2}and integrating dxdy.

the dy limits would then be

[when z=1] y

^{2}= 1 - x

^{2}

y = sqrt[1-x

^{2}]

[when z=0] y

^{2}= -x

^{2}

y = sqrt[-x

^{2}] = xi i=complex number

the dx limits would then be

[when z=1] x

^{2}= 1 - y

^{2}

x = sqrt[1-y

^{2}]

[when z=0] x

^{2}= -y

^{2}

x = sqrt[-y

^{2}] = yi i=complex number

but this seems like a dead end

any suggestions would be helpful

ok just tried another thing: the lower limits for dx and dz being zero taking out the complex numbers.

this gives: after dy=> 2y dx = 2sqrt(1-x

^{2})dx = x*sqrt(1-x

^{2}) + sin

^{-1}x

i then put in the limits 1 an 0 and got pi/2

still dont know if this is right though

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