# Thread: Cylindrical co-ordinates, triple integration

1. ## Cylindrical co-ordinates, triple integration

The integration i can do but i have no idea how you get the limits for to integrate by.

I have to integrate a function where E (the limits) is the region enclosed by the planes z = 0 and z = x + y + 5 and by the cylinders X^2 + y^2 = 4 and x^2 + y^2 = 9.

My take it is i first have to integrate with z between x + y + 5 and 0, but then i dont know how to get the next integrals with respect to x and y.

2. Originally Posted by adam_leeds
The integration i can do but i have no idea how you get the limits for to integrate by.

I have to integrate a function where E (the limits) is the region enclosed by the planes z = 0 and z = x + y + 5 and by the cylinders X^2 + y^2 = 4 and x^2 + y^2 = 9.

My take it is i first have to integrate with z between x + y + 5 and 0, but then i dont know how to get the next integrals with respect to x and y.

I presume that you recognize that $\displaystyle x^2+y^2= 4$ and $\displaystyle x^2+ y^2= 9$ are concentric cylinders of radii 2 and 3, repectively. when z= 0, the plane z= x+ y+ 5 becomes the line x+y= -5 which crosses the x and y axes at (0, -5) and (-5, 0). It's closest approach to the origin is where x= y: 2x= -5, x= y= -5/2. The distance from (-5/2, -5/2) to the origin is [tex]\sqrt{\frac{25}{4}+ \frac{25}{4}= \frac{5}{2}\sqrt{2}> 3. Whew! That means the top plane does not cut the lower plane inside the cylinders! That would have made the problem much harder!
So: In cylindrical coordinates, $\displaystyle \theta$ runs from 0 to $\displaystyle 2\pi$, r runs from 2 to 3, and z runs from 0 to z= x+ y+ 5= $\displaystyle r cos(\theta)+ r sin(\theta)+ 5$. That is:
$\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 2}^3\int_{z= 0}^{rcos(\theta)+ rsin(\theta)+ 5} f(r,\theta, z)r dzdrd\theta$
$\displaystyle \mathop\int\int\int\limits_{\hspace{-25pt}\text{Red}} f(r,\theta,z)dV$