# Thread: Triple Integrals In Cylindrical Coordinates.

1. ## Triple Integrals In Cylindrical Coordinates.

Using cylindrical coordinates, evaluate (OVER R) $\int \int \int \sqrt{x^2+y^2} dV$ , where the solid R is bounded by the surfaces $z=\sqrt{x^2+y^2}$ and z=5.

OK, I know how to graph these two surfaces, But how in the earth can I find the limits of the integration?
the " $dz$" limits: $5 \rightarrow r$
but what about " $d\theta$" and " $dr$" ??!

2. Originally Posted by TWiX
Using cylindrical coordinates, evaluate (OVER R) $\int \int \int \sqrt{x^2+y^2} dV$ , where the solid R is bounded by the surfaces $z=\sqrt{x^2+y^2}$ and z=5.

OK, I know how to graph these two surfaces, But how in the earth can I find the limits of the integration?
the " $dz$" limits: $5 \rightarrow r$
but what about " $d\theta$" and " $dr$" ??!
the inside integral $\int_r^5 ( \cdot) dz$

The two surfaces intersect giving a circle of radius 5. This is the region for the outer two integrals.

3. Originally Posted by Danny
the inside integral $\int_r^5 ( \cdot) dz$

The two surfaces intersect giving a circle of radius 5. This is the region for the outer two integrals.
Nyahahahaa
I got it
But there is a little problem
why is it from r to 5 ?
why not from 5 to r ?

4. Originally Posted by TWiX
Nyahahahaa
I got it
But there is a little problem
why is it from r to 5 ?
why not from 5 to r ?
The cone ( $z=r$) is below the plane ( $z=5$).

5. Originally Posted by Danny
The cone ( $z=r$) is below the plane ( $z=5$).
hmmmm
I did not memorize the shapes
THANK YOU

6. Originally Posted by TWiX
Nyahahahaa
I got it
But there is a little problem
why is it from r to 5 ?
why not from 5 to r ?
Originally Posted by TWiX
hmmmm
I did not memorize the shapes
THANK YOU
Well, you don't have to memorize the shapes. At (0,0), z= 0 for the cone and z= 5 for the plane. 5> 0.

Also it is neither "from r to 5" nor "from 5 to r". It is "r ranges from 0 to 5".