Results 1 to 3 of 3

Math Help - Polar Integration Question

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    81

    Polar Integration Question

    Hi all,

    The question -

    Calculate V = \int \int_R \sin{\sqrt{x^2 + y^2}} dA

    R is the region inside the circle x^2 + y^2 = 4\pi^2

    My attempt (converting to polar) - V = \int \int_R z dA

     z = \sin{\sqrt{x^2 + y^2}}

    Assuming we can find a quarter of the circle, than multiply by 4 for the total area...

    V = 4 \int^{\frac{\pi}{2}}_0 \int^{2\pi}_0 \sin{r} r dr d\theta

     = 4 \int^{\frac{\pi}{2}}_0 [\sin{r} - r\cos{r}]^{2\pi}_{0} d\theta

     = 4 \int^{\frac{\pi}{2}}_0 [\sin{2\pi} - 2\pi\cos{2\pi}] d\theta

    Here I'm stuck and quite frankly it's looking a lot more complicated than I imagine it should, can someone please point out the error in my working?

    On the off chance I am correct so far, can I treat those 2\pi sections as radians? I don't believe I can as that wasn't the context they were used originally, so what would I do?

    Thanks for your time folks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Peleus View Post
    Calculate V = \int \int_R \sin{\sqrt{x^2 + y^2}} dA

    R is the region inside the circle x^2 + y^2 = 4\pi^2

    My attempt (converting to polar) - V = \int \int_R z dA

     z = \sin{\sqrt{x^2 + y^2}}

    Assuming we can find a quarter of the circle, than multiply by 4 for the total area...

    V = 4 \int^{\frac{\pi}{2}}_0 \int^{2\pi}_0 \sin{r} r dr d\theta

     = 4 \int^{\frac{\pi}{2}}_0 [\sin{r} - r\cos{r}]^{2\pi}_{0} d\theta

     = 4 \int^{\frac{\pi}{2}}_0 [\sin{2\pi} - 2\pi\cos{2\pi}] d\theta

    Here I'm stuck and quite frankly it's looking a lot more complicated than I imagine it should, can someone please point out the error in my working?

    On the off chance I am correct so far, can I treat those 2\pi sections as radians? I don't believe I can as that wasn't the context they were used originally, so what would I do?
    Yes, you are correct so far, and 2\pi means 2\pi whatever context it appears in. So the next step is to say that \sin2\pi=0 and \cos2\pi=1, and you are just integrating the constant -2\pi with respect to \theta. (If you're worried that the answer comes out negative, think what the graph of the function looks like, and you'll see that the answer should indeed be negative.)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    Posts
    81
    Quote Originally Posted by Opalg View Post
    Yes, you are correct so far, and 2\pi means 2\pi whatever context it appears in. So the next step is to say that \sin2\pi=0 and \cos2\pi=1, and you are just integrating the constant -2\pi with respect to \theta. (If you're worried that the answer comes out negative, think what the graph of the function looks like, and you'll see that the answer should indeed be negative.)
    Firstly, thank you for your fast reply.

    Obviously it's a big confidence boost to know I'm getting something right, it's been a hell of a night

    In regards to the 2\pi however, isn't it used in the context of a distance (the radius) rather than radian's for the sin and cos?

    So wouldn't \sin2\pi=0 and \cos2\pi=1 only be correct if it was in fact radians, which from my understanding it is not? For some reason I'm thinking it should be something like \sin{6.28...} = 0.1094.....

    I'm sure I'm wrong, but it's always important to find out why your wrong, rather than simply getting the right answer
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] integration...should i use polar transformation?
    Posted in the Calculus Forum
    Replies: 13
    Last Post: October 21st 2010, 09:19 AM
  2. Polar Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 4th 2010, 06:54 PM
  3. Replies: 3
    Last Post: March 28th 2009, 06:27 AM
  4. Polar Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2008, 12:18 PM
  5. Polar Coordinate Integration
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 16th 2006, 07:51 AM

Search Tags


/mathhelpforum @mathhelpforum