Results 1 to 7 of 7

Math Help - triple integral problem

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ll0124 View Post
    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 1\leq z\leq 5-y but in cylindrical coordinates, y=r\sin\theta...

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

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    10
    i know... i wasn't sure when to change that.

    thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    10
    okay i am getting 27pi now.. which is at least close.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    10
    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..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ll0124 View Post
    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 \int_1^{5-r\sin\theta}r\,dz, we have r\left[(5-r\sin\theta)-1\right]=4r-r^2\sin\theta.

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

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    10
    perfect sense, thank you so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Triple Integral Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 7th 2011, 12:52 AM
  2. another triple integral problem..
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 3rd 2010, 03:04 AM
  3. Triple integral problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 3rd 2009, 05:01 PM
  4. Triple integral problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 24th 2009, 09:12 AM
  5. Another triple Integral problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 6th 2008, 04:45 PM

Search Tags


/mathhelpforum @mathhelpforum