help in integration?

a a

j j x dxdy/(sqrt(x^2+y^2)

0 y

j-integrate

can you show me how to solve this (by changing the region of j)

4 Answers

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  • Luiz S
    Lv 7
    1 decade ago
    Favorite Answer

    You were creative! ! !

    a a

    ∫∫x dxdy/(x²+y²)^¹/2

    0 y

    You solve first:

    a

    ∫x / [(x²+y²)^¹/2] dx

    y

    u = x²+y²

    du = 2x dx

    du/2 = x dx

    =>

    (1/2)*∫1/[u^¹/2] du =(1/2)*∫u^-¹/2 du = (1/2)(2/1)*u^¹/2

    =>

    ∫x / [(x²+y²)^¹/2] dx = (x²+y²)^¹/2 + c

    =>

    a ................ ............... ............... a

    ∫x / [(x²+y²)^¹/2] dx = (x²+y²)^¹/2 . |

    y ................. .................. ............ y

    a

    ∫x / [(x²+y²)^¹/2] dx = (a²+y²)^¹/2 - (y²+y²)^¹/2

    y

    =>

    a

    ∫[(a²+y²)^¹/2 - y*2^¹/2] dy

    0

    I think you dont need to change the region, only solve the integration as it were indefinied.

    Do the rest...

  • 1 decade ago

    i'm a little rusty in this, had solved this type of question 2 yrs ago. i'll explain the base, but i can't get any further. maybe someone else can understand & take t to the next step..

    there's two integrations: first one is of y, from 0 to a.

    next one is of x, from y to a.

    (we're assuming this, 'coz if y went from y to a, all hell would break lose.)

    take a x-y axis, and picture a square of sides a *a

    now imagine a diagonal passing thru points (0,0) and (a,a).

    our region of integration here is the lower right triangle ,that is, with (x,y) co-ordinate points being (0,0), (a,0) and (a,a) (you better draw this on paper!)

    make some arrows inside the triangle, going from left to right. this denotes x being operative variable.

    now comes the region-change. in the problem, y is inside fixed limits, & x is operating in variable. we want to make y operate in terma of x, and x should be fixed!

    that's easy: our triangle lies within x=0 to x=a.

    taking the diagonal for y, y goes from y=0 upto only y=x (make the arrows in the triangle go upwards)

    so now, x :: 0 to a

    y :: 0 to x

    so now our question becomes:

    [integral of x from 0 to a] * [integral of y from 0 to x] * x / sqrt(x^2+y^2) dx dy.

    do y-integration only first, treating x as constant:

    (note: arcsin means sine inverse)

    integral of k dy/ sqrt (k^2 + y^2) = k * -arcsin(y/k)

    >> -x * arcsin (y/x)

    limits: put y, from 0 to x

    >>[ -x * arcsin(x/x) ] - [ - x* arcsin(0/x) ] ..... (remember the original x is still treated as constant here)

    >> - x* arcsin(1) = -x * pi/2 (or -90 degrees)

    Now, put this back, into the next (fixed) integral:

    integral ( -x * pi/2 * dx) from x = 0 to a

    >> [-pi/2 * 1/2 * x^2 ] 0 to a

    >> -a^2 * pi / 4

    is that the defintive answer?? - a^2 * pi/4 ?? i dunno... it's been a while since i did this stuff!

    check out http://www.alcyone.com/max/reference/maths/integra...

    best of luck.

  • 1 decade ago

    i'm not getting your equation. try to write it differently. i don't understand the 2 a's or what the 0y is doing there

  • 1 decade ago

    That is not any maths.What are you mean by the 0y...???

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