Triple integral to find volume

**Calculate the volume bounded by the surface 9 + x + y = z , z = 1 , x^2 +y^2 = 4?**

i try to convert the range into cylindrical coordinate first..

i know the range for phi and r..

i just dont know the range of z..

where i try z is from 0 to 9 + r (dont know whether it is correct or not..)

Re: Triple integral to find volume

Quote:

Originally Posted by

**nameck** **Calculate the volume bounded by the surface 9 + x + y = z , z = 1 , x^2 +y^2 = 4?**

i try to convert the range into cylindrical coordinate first..

i know the range for phi and r..

i just dont know the range of z..

where i try z is from 0 to 9 + r (dont know whether it is correct or not..)

This surface is formed from the cylinder $\displaystyle x^2+y^2=4$ by slicing it off at the bottom by the plane z=1, and at the top by the plane z= 9+x+y. So the lower limit for z is 1 (not 0), and the upper limit is $\displaystyle 9+x+y$, which you have to write as $\displaystyle 9+r(\cos\theta+\sin\theta)$.