Problem for the Calculus MASTERS

[jmcoreymv]

I just had this problem on a midterm, and I came up with some whacked out answer and I wanted to see if someone else could solve it

integral of (y+3x)^(1/3) * (x+y)^(1/2) dydx with the y integral limits from 0 to 2-x and the x integral limits from 0 to 2. I tried solving it using UV substitution.

 

[Ausm]

Originally posted by: jmcoreymv
I just had this problem on a midterm, and I came up with some whacked out answer and I wanted to see if someone else could solve it

integral of (y+3x)^(1/3) * (x+y)^(1/2) dydx with the y integral limits from 0 to 2-x and the x integral limits from 0 to 2. I tried solving it using UV substitution.

I would help but do to alcohol consumption my calculations might be a little off

Sysadmin

 

[jmcoreymv]

^

 

[makessense]

can't you use your ti-86 or ti-89 to solve it?

 

[jmcoreymv]

It took several minutes and spit out a long decimal number. Im looking for the process.

 

[jmcoreymv]

The secret must be known....

 

[AgaBoogaBoo]

lol

 

[attahoova]

try u sub twice, int by parts twice


it has the look of a "twice" integral

 

[jmcoreymv]

Im pretty positive its supposed to be solved using uv substitution. I just always have trouble determine the new area for the limits of the u and v variables. I believe i set u = y+3x and v = x+y and went from there and ended up getting a shape to enclose with v=2 u=v and v=-u/3. My jacobian also came out to 1 which seemed very nice.

 

[jmcoreymv]

Bump for an awesome poster...

 

[Evadman]

Originally posted by: jmcoreymv
Bump for an awesome poster...

You called?

 

[jessicak]

just give him an answer so he'll stop making me bump this thread. Thanks

 

[jmcoreymv]

Turns out the crazy ass answer I got was right and the professor said he just made a mistake when he made the problem and made it way too hard.