Using Jacobian to determine area

In summary: What do you mean by "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"?I don't know, I guess you're saying that just multiplying the area of the rectangle times the jacobian would work just as well?No, I'm saying that 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.No, I'm saying that the area of the rectangle is (7-1)(5-2)= 6(3)= 18 so the area of the transformed region would be 18
  • #1
snoggerT
186
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?
 
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  • #2
Well, what did you do? What is the Jacobian? And what answer did you get?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 

FAQ: Using Jacobian to determine area

What is the Jacobian method used for?

The Jacobian method is used in mathematics and physics to determine the area of a region in a multi-dimensional space or the volume of a solid object. It is a useful tool for calculating integrals and solving differential equations.

How does the Jacobian method work?

The Jacobian method involves calculating the determinant of the Jacobian matrix, which is a matrix of partial derivatives. This matrix represents how the coordinates of a point in one space (such as Cartesian coordinates) are related to the coordinates in another space (such as polar coordinates). By using this matrix and its determinant, the area or volume of the region can be determined.

When should the Jacobian method be used?

The Jacobian method should be used when the region or object in question cannot be easily described or calculated using traditional methods. It is particularly useful for complex shapes or in higher dimensions.

What are the advantages of using the Jacobian method?

The Jacobian method allows for the calculation of areas and volumes in a more general and flexible way than traditional methods. It also allows for the transformation of coordinates, making it useful in a variety of mathematical applications.

Are there any limitations to using the Jacobian method?

While the Jacobian method is a powerful tool, it does have some limitations. It can only be used for regions or objects that can be described using a coordinate system, and it may not always be the most efficient method for calculating areas or volumes. Additionally, it requires a good understanding of multi-dimensional calculus and linear algebra to use effectively.

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