Changing the variable in multiple integrals

kidsmoker
Messages
85
Reaction score
0

Homework Statement



Evaluate

\int\int(x-y)^2sin^2(x+y)dxdy

taken over a square with successive vertices (pi,0), (2pi,pi), (pi,2pi), (0,pi).


Homework Equations



I = \int\int_{K} f(x,y)dxdy = \int\int_{K'} g(u,v)*J*dudv

where J is the Jacobian.

The Attempt at a Solution



Okay so I've just been learning this for the first time, so I may be doing it completely wrong!

I used the transformations u=x-y, v=x+y which give the Jacobian as 2.

Now i wasn't sure how to get the new limits for the integrals. What I did was apply the transformation above to give new vertices:

(pi,0) -> (pi,pi)
(0,pi) -> (-pi,pi)
(pi,2pi) -> (-pi,3pi)
(2pi,pi) -> (pi,3pi)

This gives a simple rectangle, so then i just wrote

I = 2*\int^{3\pi}_{\pi}\int^{\pi}_{-\pi}u^2sin^2(v)dudv = \frac{4\pi^{4}}{3}.

I wish this was right, but I've a feeling it's not :-(

Any help greatly appreciated!
 
Last edited:
Physics news on Phys.org
The rectangle looks ok. But haven't you got the jacobian factor upside down?
 
Ah yeah, should be 1/2. Other than that though does my method look correct?

Thanks.
 
kidsmoker said:
Ah yeah, should be 1/2. Other than that though does my method look correct?

Thanks.

Looks ok to me.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top