Find the volume of the region that lies inside z=x^2+y^2 and below the plane z=16.

Thank you very much.

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- Jul 31st 2007, 07:14 PMkittycatintegral calculus
Find the volume of the region that lies inside z=x^2+y^2 and below the plane z=16.

Thank you very much. - Jul 31st 2007, 07:18 PMThePerfectHacker
- Jul 31st 2007, 07:37 PMkittycat
Hi,

Thank you very much for your reply.

Could you please explain to me why x^2 + y^2 -16 ? Thanks - Jul 31st 2007, 08:19 PMThePerfectHacker
- Jul 31st 2007, 09:01 PMkittycat
hi Perfecthacker,

I know

V=double integral of 16 dA is the volume under z=16

V= double integral of x^2+y^2 dA is the volume under z= x^2 +y^2

therefore the desired volume is the difference between them.

But what I don't understand is

why the desired volume is the difference between these two double integrals? Could you please explain to me? Thank you very much. - Jul 31st 2007, 09:10 PMtukeywilliams
because you want to find the volume that is inside the cone $\displaystyle z = x^2 + y^2 $ and below the plane $\displaystyle z = 16 $.

$\displaystyle \iint_{D} 16 \ dA $ is the volume**under**$\displaystyle z = 16 $. $\displaystyle \iint_{D} x^2+y^2 \; dA $ is the volume**under**$\displaystyle z = x^2 + y^2 $. $\displaystyle z = 16 $ is above that part of the cone. So we want to find the volume that is bounded by the plane and the cone. In other words, $\displaystyle z = 16 $ is 'higher' than $\displaystyle z = x^2+y^2 $. So when you draw in $\displaystyle z = 16 $ there is still some part of the cone left underneath.

http://tutorial.math.lamar.edu/AllBr...s/image005.gif - Jul 31st 2007, 09:32 PMkittycat
Thank you very much turkeywilliam. :):)