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    Math help please ! - double integral

    Not sure if i have done this question right i would appreciate anyone who could solve it for me who is good at math and let me know what answer you get !.....
    and yes i have used wolframalpha but i am a bit sceptic on wether i have inputed it correctly etc.

    I get a negative exponential btw

    12 Comments

    Original Poster

    http://img25.imageshack.us/img25/4567/84914400.jpg

    My head just exploded.

    Original Poster

    scott1295;6762041

    My head just exploded.



    haha , mechanical engineering believe it or not !:w00t: :thinking:

    I worked it out to be 5.

    I didn't use wolframalpha, I guessed.

    :P

    LM290;6762049

    haha , mechanical engineering believe it or not !:w00t: :thinking:



    Do you not treat the two integration bits seperately and then multiple the answers from the 2 collecting like terms etc.

    Original Poster

    scott1295;6762091

    Do you not treat the two integration bits seperately and then multiple … Do you not treat the two integration bits seperately and then multiple the answers from the 2 collecting like terms etc.



    yup correct, however in the lecture the examples give gave simple answers lik 28/3 etc. for this i get ( - 1673895616/ 165 ) :w00t: so i am a bit unsure if i am doin it correct, my mate gets totally different answers from me so would like another outside opinion

    worked longhand get approx -million like your calc

    I got 8008135 (or should that be moobies) :oops:

    where in the world (work place) would you need to use/have that info :thinking:

    i remember having an argument with my maths teacher many moons about algerbra & i said to him i'd never need to use it & i was right !

    Basically you integrate wrt y first and substitute the 2x^3 and 3x limits for y. You then integrate resulting expression wrt x and then put in the 2 , 4 limits that gives result

    hth

    hughwp;6763561

    Basically you integrate wrt y first and substitute the 2x^3 and 3x limits … Basically you integrate wrt y first and substitute the 2x^3 and 3x limits for y. You then integrate resulting expression wrt x and then put in the 2 , 4 limits that gives resulthth



    yeah :thinking: :?

    Original Poster

    hughwp;6763561

    Basically you integrate wrt y first and substitute the 2x^3 and 3x limits … Basically you integrate wrt y first and substitute the 2x^3 and 3x limits for y. You then integrate resulting expression wrt x and then put in the 2 , 4 limits that gives resulthth


    done that but its the value i want incase my basic maths fails me like it usually does :roll:, it looks like somone else tried it and got the same as me so im happy with that
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