double integrals

• Apr 6th 2007, 10:20 PM
elbarto
double integrals
Hi,
Could somebody please help me with my assignment. I am up to the last question which involves multiple intergration. my problem with question a is that when i try to intergrate it with respect to x i get stuck after just a few lines. i have been useing integration by parts letting u=xy and dv=(1+x^4)^-0.5. could somebody please give me an idea on what is the best way to approach this (i dont think we are allowed to use intergrals for this problem). As for question b, i am yet to attempt the question but from inspection i dont entirely understand what dA is spose to represent. I assume we are spose to integrate over the contour x^2+Y^2=1 but why whould this be writen as x^2+Y^2 =< 1?

Thank you for any help you can give me, it will be greatly appreciated.
Regards Elbarto
• Apr 6th 2007, 10:32 PM
CaptainBlack
Quote:

Originally Posted by elbarto
Hi,
Could somebody please help me with my assignment. I am up to the last question which involves multiple intergration. my problem with question a is that when i try to intergrate it with respect to x i get stuck after just a few lines. i have been useing integration by parts letting u=xy and dv=(1+x^4)^-0.5. could somebody please give me an idea on what is the best way to approach this (i dont think we are allowed to use intergrals for this problem). As for question b, i am yet to attempt the question but from inspection i dont entirely understand what dA is spose to represent. I assume we are spose to integrate over the contour x^2+Y^2=1 but why whould this be writen as x^2+Y^2 =< 1?

Thank you for any help you can give me, it will be greatly appreciated.
Regards Elbarto

Integral_{(x,y) in T} xy/(1+x^4) dA = Integral_{x=0 to 1} (Integral_{y=0 to x} xy/(1+x^4) dy) dx

where T is the interior of the triangle with vertices (0,0), (1,0), (1,1), and dA is the area element
which in cartesians is dx dy (or dy dx whichever is more convienient)

Does that help?

RonL
• Apr 7th 2007, 05:57 PM
ThePerfectHacker
..
• Apr 7th 2007, 06:22 PM
ThePerfectHacker
The last integral is remcommended to be done in polar coordinates.

Then,
x^2+y^2=r^2

Thus,

INT (r=0 to r=1) INT (@=0 to @=2pi) r ln r^2 dr d@

But this can easily be done under the variable substitution,
u=r^2
• Apr 9th 2007, 12:14 AM
elbarto
Thanks for the help so far,
CaptainBlack i took what you said about integrating which eva variable is easier first and decided to integrate wrt y first. I think i have made a mistake with my limits as my answer did not match the answer that theperfecthacker provided. i tried useing substition as substition and integration by parts are two methods that i have been useing in my coarse (still just picking them up). Theperfecthacker, when you itegrated the first question did you arange it all just by inspection knowing that the integral will be simplified or is there a technique u used to simplify it.

Regards Elbarto
• Apr 9th 2007, 12:52 AM
Jhevon
Quote:

Originally Posted by elbarto
Thanks for the help so far,
CaptainBlack i took what you said about integrating which eva variable is easier first and decided to integrate wrt y first. I think i have made a mistake with my limits as my answer did not match the answer that theperfecthacker provided. i tried useing substition as substition and integration by parts are two methods that i have been useing in my coarse (still just picking them up). Theperfecthacker, when you itegrated the first question did you arange it all just by inspection knowing that the integral will be simplified or is there a technique u used to simplify it.

Regards Elbarto