Using Jacobian to determine area

  • Thread starter Thread starter snoggerT
  • Start date Start date
  • Tags Tags
    Area Jacobian
snoggerT
Messages
183
Reaction score
0
Let [phi](u,v)=(3u+v,u-2v). Use the Jacobian to determine the area of [phi]R for:

R=[2,5]X[1,7]




The Attempt at a Solution



- I'm really not sure why I keep getting the wrong answer on this problem. the problem gives you two R's to solve for and I got the right answer for the first one (R=[0,3]X[0,5]), but I'm not getting the right answer for the 2nd R. I would think you would solve the problems the exact same way, but I'm not sure since I can't get the right answer. Would R being changed affect how you work the problem?
 
Physics news on Phys.org
Well, what did you do? What is the Jacobian? And what answer did you get?
 
HallsofIvy said:
Well, what did you do? What is the Jacobian? And what answer did you get?

- my jacobian was |-7|. I set the limits on my outer integral (for dv) from 1 > 7 and my inner integral from 2 > 5. I integrated with respect to u first and then integrated with respect to v. I kept getting a very large negative number, but the answer is positive and not that big.
 
How could you possibly have gotten negative answer? The Jacobian is, as you say, 7, and the area of the rectangle is (7-1)(5-2)= 6(3)= 18 so the area of the transformed region would be 18(7)= 126.
 
HallsofIvy said:
How could you possibly have gotten negative answer? The Jacobian is, as you say, 7, and the area of the rectangle is (7-1)(5-2)= 6(3)= 18 so the area of the transformed region would be 18(7)= 126.

- So you don't use integration on this problem? You can just multiply the area of the rectangle times the jacobian?
 
that's the whole point of the problem! No you don't need to use integration to find the area of a rectangle and the point of the Jacobian is that it changes the area of one region to the area of the other.
 
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