# 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

• Apr 9th 2007, 04:20 AM
elbarto
Could it be that i am intergrating over the wrong area? If 0<y<x and 0<x<1 that means im integrating over the area bound between y=x and the x axis right???? will that be the same as integrating over the area bound by y=x and the y axis? If not is there anything i can do to correct this useing my method or do i need to use the method that the perfect hacker has used.

Thanks, Elbarto
• Apr 10th 2007, 04:58 PM
elbarto
Just wanted to say thanks to you all for all the help you have given me with this problem. i realised that i made a mistake in my first set of limits. instead of 0<y<x it should be x<y<1. The answer i get is 0.1097, i think that is the number theperfecthacker ment to type in his previous post.

I integgrated the function over the area of the rectangle 0<x<1 and 0<y<1 to give me a value of 0.1963 then i subtracted the value 0.0866 that i had found earlier to give me the final answer of 0.1097.

Thanks Again
Elbarto