# integral calculus

• July 31st 2007, 07:14 PM
kittycat
integral 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.
• July 31st 2007, 07:18 PM
ThePerfectHacker
Quote:

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

Thank you very much.

$\iiint_V \ dV = \iint_R \int_{x^2+y^2}^{16} dz \ dA = \iint_R x^2+y^2-16 \ dA$

$\int_0^{2\pi} \int_0^4 (r^2-14)r\ dr d\theta$
• July 31st 2007, 07:37 PM
kittycat
Hi,

Could you please explain to me why x^2 + y^2 -16 ? Thanks
• July 31st 2007, 08:19 PM
ThePerfectHacker
Quote:

Originally Posted by kittycat
Hi,

Could you please explain to me why x^2 + y^2 -16 ? Thanks

It should be,
$16-x^2-y^2$
• July 31st 2007, 09:01 PM
kittycat
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.
• July 31st 2007, 09:10 PM
tukeywilliams
because you want to find the volume that is inside the cone $z = x^2 + y^2$ and below the plane $z = 16$.

$\iint_{D} 16 \ dA$ is the volume under $z = 16$. $\iint_{D} x^2+y^2 \; dA$ is the volume under $z = x^2 + y^2$. $z = 16$ 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, $z = 16$ is 'higher' than $z = x^2+y^2$. So when you draw in $z = 16$ there is still some part of the cone left underneath.

http://tutorial.math.lamar.edu/AllBr...s/image005.gif
• July 31st 2007, 09:32 PM
kittycat
Thank you very much turkeywilliam. :):)