# Thread: integral calculus

1. ## 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.

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

$\displaystyle \int_0^{2\pi} \int_0^4 (r^2-14)r\ dr d\theta$

3. Hi,

Thank you very much for your reply.

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

4. Originally Posted by kittycat
Hi,

Thank you very much for your reply.

Could you please explain to me why x^2 + y^2 -16 ? Thanks
It should be,
$\displaystyle 16-x^2-y^2$

5. 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.

6. 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.

7. Thank you very much turkeywilliam.