Surface Integral Homework: Solving with Variable Substitution

NT123
Messages
26
Reaction score
0

Homework Statement

I am to use a substitution of variables u = x, v = x + 2y to evaluate the surface integral

int(0,1/2)int(0,1-y) exp(x/(x+y))dxdy

where int(a,b) means integral sign with lower limit a and upper limit b.



Homework Equations





The Attempt at a Solution

I used the substitution and calculated the Jacobian to be 1/2. However, in the new variables, the integral is now

int(0,1/2)int(0,1+u-v)exp(u/v)dudv

Both variables appearing in one of the integrals doesn't seem right to me - what has gone wrong with the calculation?
 
Physics news on Phys.org
NT123 said:
int(0,1/2)int(0,1+u-v)exp(u/v)dudv

Both variables appearing in one of the integrals doesn't seem right to me - what has gone wrong with the calculation?

Are you sure you have all factors of 2 correct? E.g. in the denominator of the exponential function, and the integration limit for the inner integral?

It is not a problem that the outer integration variable is part of the inner integral. The opposite however, doesn't make sense.

Torquil
 
I get different things for both integrand and limits of integration. Putting x= u into v= x+ 2y gives v= u+ 2y so that y= (v- u)/2. Then x+ y= (v- u)/2+ 2u/2= (u+ v)/2. x/(x+y)= 2u/(u+v), not u/v.

In the x,y plane, the region to be integrated is the quadrilateral with sides x= 0, y= 0, y= 1/2 and x+ y= 1.

With x= 0, u= 0, v= 2y or just u= 0. With y= 0, u= x, v= x or v= u. With y= 1/2, u= x, v= x+1 or v= u+1. Adding u= x and v= x+ 2y gives u+v= 2x+ 2y= 2(x+ y) so with x+y= 1, u+v= 2.

In the xy, plane, the region is a quadrilateral bounded by u= 0, v= u, v=u+1, and u+ v= 2. Now, whether you integrate with respect to u or v first, you are going to have to break that into two pairs of integrals. If you integrate with respect to v first, you will have to have one pair of integrals with u going from 0 to 1/2 and then another with u going from 1/2 to 1. If you integrate with respect to u first, you will have to have a pair of integrals with v going from 0 to 1 and then from 1 to 3/2.
 
Sorry, I made a typo, the integral was supposed to be int(0,1/2)int(0,1-2y)exp(x/(x+2y)), hence the substitution.
Thanks for the help though!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top