Results 1 to 6 of 6

Math Help - Surface integral with exponentials

  1. #1
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182

    Surface integral with exponentials

    Hi, I've been looking at surface integrals, and I'm having a problem with all the exponential functions I come across. In a single variable case:

    \int_{a}^b e^{2x+1} dx

    I would just take this as:

    \int_{a}^b e^{2x+1} dx = \int_{a}^b e^{u} du = [\frac{e^u}{f'(u)}]

    and then take the integral within the new limits given by du = 2dx.

    An example of a question with exponentials in a surface integral that I'm stuck on is, evaluate:

    \int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx

    How do I go about such a thing? I know how to do some basic surface integrals, with indicies/trig functions etc, over rectangular and non-rectangular domains BTW, but this exponential stuff has got me beat. Thanks

    P.S. how can I get the brackets properly round the fraction with Latex? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Greengoblin View Post
    Hi, I've been looking at surface integrals, and I'm having a problem with all the exponential functions I come across. In a single variable case:

    \int_{a}^b e^{2x+1} dx

    I would just take this as:

    \int_{a}^b e^{2x+1} dx = \int_{a}^b e^{u} du = [\frac{e^u}{f'(u)}]

    and then take the integral within the new limits given by du = 2dx.

    An example of a question with exponentials in a surface integral that I'm stuck on is, evaluate:

    \int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx

    How do I go about such a thing? I know how to do some basic surface integrals, with indicies/trig functions etc, over rectangular and non-rectangular domains BTW, but this exponential stuff has got me beat. Thanks

    P.S. how can I get the brackets properly round the fraction with Latex? Thanks
    you evaluate it the same way. when you're dealing with dy, treat x as a constant, when dealing with dx treat y as a constant. that is, when doing \int e^{x + y}~dx, it might as well be \int e^{x + 1}~dx, you'd treat the y as any other constant, here i used 1, but it could be anything in your head. the same rules apply

    can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182
    thanks, i'll try:

    u=(x+y), \frac{du}{dy} = 1, du=dy

    \int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx = \int_{0}^2  [\frac{e^{u}}{du/dy}]_{x}^{2x} dx = \int_{0}^2 [\frac{e^{u}}{1}]_{x}^{2x} dx =\int_{0}^2 [e^{x+y}]_{x}^{2x} dx

    = \int_{0}^2 [e^{3x} - e{2x}]dx = [\frac{e^{3x}}{3} - \frac{e^{2x}}{2}]_{0}^2 = \frac{e^6}{3} - \frac{e^4}{2} - \frac{1}{3} + \frac{1}{2}

    ...and then calculate. d'ya reckon thats ok? Thanks

    Sorry about the brackets I don't know how to get them around the fraction properly - for everything I try it has a latex error. Also I spose alot of the 'd's should be the wigly partial derivative sign, but I don't know how to get that either.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Greengoblin View Post
    thanks, i'll try:

    u=(x+y), \frac{du}{dy} = 1, du=dy

    \int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx = \int_{0}^2  [\frac{e^{u}}{du/dy}]_{x}^{2x} dx = \int_{0}^2 [\frac{e^{u}}{1}]_{x}^{2x} dx =\int_{0}^2 [e^{x+y}]_{x}^{2x} dx

    = \int_{0}^2 [e^{3x} - e{2x}]dx = [\frac{e^{3x}}{3} - \frac{e^{2x}}{2}]_{0}^2 = \frac{e^6}{3} - \frac{e^4}{2} - \frac{1}{3} + \frac{1}{2}

    ...and then calculate. d'ya reckon thats ok? Thanks
    you got the idea. i just think you'd want to simplify more. maybe get everything in one fraction or so

    Sorry about the brackets I don't know how to get them around the fraction properly - for everything I try it has a latex error.
    use \left[ \right]

    this works in general to wrap brackets around anything. so you could do \left( \right) or \left \{ \right \} etc

    Also I spose alot of the 'd's should be the wigly partial derivative sign, but I don't know how to get that either.
    type \partial to get \partial
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182
    Thanks alot, I spose I could get everthing as a fraction over 6, but it was getting hard to decipher the latex I've already written . I've got a couple more of these questions too, but I'll have another go at them, and I'll post my working back here later. Cheers!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Greengoblin View Post
    Thanks alot, I spose I could get everthing as a fraction over 6, but it was getting hard to decipher the latex I've already written . I've got a couple more of these questions too, but I'll have another go at them, and I'll post my working back here later. Cheers!
    ok, that's good. good luck.

    (you could write it out on paper first and then translate to LaTex)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Integral
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 3rd 2011, 04:28 AM
  2. Replies: 11
    Last Post: March 7th 2011, 08:59 AM
  3. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 06:11 PM
  4. Surface integral.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 19th 2010, 04:22 AM
  5. Surface Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 22nd 2009, 07:46 PM

Search Tags


/mathhelpforum @mathhelpforum