1. ## triple integral problem

use a triple integral to find the volume of the given solid

the solid enclosed by the cylinder x^2+y^2=9 and the planes y+z=5 and z=1

i attempted to draw a picture..

and i ended up with the following integral:
(i don't know how to write math symbols on here)
int from 0 to 2pi int from 0 to 3 int from 1 to y-5 of dzrdrdtheta

i'm positive that's wrong.
since it's a cylinder, i figured polar seemed like a good idea.. and i checked the answer in the back and it's 36pi. the pi leads me to think i should use polar.
so my first limits are 0 and 2pi followed by 0 and 3, since 3 is the radius of the cylinder.
it is after that that i get very confused.
i think i should be integrating from 1 to y=5 because those are the two planes when set equal to z...
but i'm not sure.
and i never know what to actually integrate, so i just had dzrdrdtheta...

help?
the answer is 36pi, i just don't know how to get there.

2. Originally Posted by ll0124
use a triple integral to find the volume of the given solid

the solid enclosed by the cylinder x^2+y^2=9 and the planes y+z=5 and z=1

i attempted to draw a picture..

and i ended up with the following integral:
(i don't know how to write math symbols on here)
int from 0 to 2pi int from 0 to 3 int from 1 to y-5 of dzrdrdtheta

i'm positive that's wrong.
since it's a cylinder, i figured polar seemed like a good idea.. and i checked the answer in the back and it's 36pi. the pi leads me to think i should use polar.
so my first limits are 0 and 2pi followed by 0 and 3, since 3 is the radius of the cylinder.
it is after that that i get very confused.
i think i should be integrating from 1 to y=5 because those are the two planes when set equal to z...
but i'm not sure.
and i never know what to actually integrate, so i just had dzrdrdtheta...

help?
the answer is 36pi, i just don't know how to get there.
Note that $\displaystyle 1\leq z\leq 5-y$ but in cylindrical coordinates, $\displaystyle y=r\sin\theta$...

Thus, the triple integral would be $\displaystyle \int_0^{2\pi}\int_0^3\int_1^{5-r\sin\theta}r\,dz\,dr\,d\theta$

Does this make sense?

3. i know... i wasn't sure when to change that.

thank you

4. okay i am getting 27pi now.. which is at least close.

5. so i am sure there are still a lot of stupid arithmetic errors in my work that are causing this but i can't find them:

int 0 to 2pi int 0 to 3 (5r - r^2sintheta)drdtheta
= int 0 to 2 pi (5r^2/2 - (r^3)sintheta/3 from 0 to 3 dtheta
= int 0 to 2pi (45/2 - 9sintheta) dtheta
=45theta/2 + 9costheta from 0 to 2pi
=45pi + 9cos2pi - 9cos0
=45pi + 9 - 9?

i'm close..

6. Originally Posted by ll0124
so i am sure there are still a lot of stupid arithmetic errors in my work that are causing this but i can't find them:

int 0 to 2pi int 0 to 3 (5r - r^2sintheta)drdtheta
= int 0 to 2 pi (5r^2/2 - (r^3)sintheta/3 from 0 to 3 dtheta
= int 0 to 2pi (45/2 - 9sintheta) dtheta
=45theta/2 + 9costheta from 0 to 2pi
=45pi + 9cos2pi - 9cos0
=45pi + 9 - 9?

i'm close..
Note that when you integrate $\displaystyle \int_1^{5-r\sin\theta}r\,dz$, we have $\displaystyle r\left[(5-r\sin\theta)-1\right]=4r-r^2\sin\theta$.

So you then have to integrate $\displaystyle \int_0^{2\pi}\int_0^3 4r-r^2\sin\theta\,dr\,d\theta$

Does this make sense?

7. perfect sense, thank you so much!