Multivariable Calc Problem: Change of Variables in Multiple Integrals (Jacobian)

jayjay5531

New member
Joined
Nov 23, 2014
Messages
5
I just want to make sure I did this problem right. The problem was: Find ∫∫D (x + y) dV, where D is the rhombus with vertices (0, 0); (5, 0); (5/2, 5/2); and (5/2, 5/2). We were given these formulas to use for change of variables: x = 2u + 3v, and y = 2u 3v.

After sketching the region and calculating the Jacobian Determinant, I came up with the following integral:
−12 5/65/4 4u du dv (the lower limits of integration are supposed to be 0, I didn't know how to format it though). My final answer was 125/4. Somehow this feels wrong?? I guess I didn't expect it to be negative. Did I do something wrong?
 
I am a bit puzzled by "dV". The rhombus you give is a plane figure so I would have expected "dA", a differential of plane area, not "dV", a differential of volume.


In any case, the rhombus has sides given by y= x, y= x- 5, y= -x, and y= - x+ 5 If x= 2u+ 3v and y= 2u- 3v, y= x becomes 2u- 3v= 2u+ 3v so that -6v= 0 and then v= 0, y= x- 5 becomes 2u- 3v= 2y+ 3v+ 5 so v= -5/6, y= -x becomes 2u- 3v= -2u- 3v so 4u= 0 so u= 0, and y= -x+ 5 becomes 2u- 3v= -2u- 3v+ 5 or u= 5/4. That is, the rhombus, in the xy-plane, becomes a square, with side parallel to the axes, in the uv-plane.

With x= 2u+3v and y= 2u- 3v, then x+ y= 4u. The limits of integration are u= 0 to u= 5/4 and v= 0 to v= -5/6.

The Jacobian determinant is xuyuxvvy\displaystyle \left|\left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial v}{\partial y}\end{array}\right|\right|=2233=66=12\displaystyle = \left|\left|\begin{array}{cc} 2 & 2 \\ 3 & -3\end{array}\right|\right|= |-6- 6|= 12.

dxdy=12dudv\displaystyle dxdy= 12 dudv so the integral becomes v=05/6u=05/44u(12dudv)=1205/605/44ududv\displaystyle \int_{v= 0}^{-5/6}\int_{u= 0}^{5/4} 4u (12 dudv)= 12\int_0^{-5/6}\int_0^{5/4} 4u dudv

Perhaps you missed the fact that the new differential is the absolute value of the Jacobian (if it were not then the integral would change sign from "dxdy" to "dydx" and we don't want that here). (I don't know how you got that the integrand is 4 rather than 4u.)
 
Last edited:
Perhaps you missed the fact that the new differential is the absolute value of the Jacobian (if it were not then the integral would change sign from "dxdy" to "dydx" and we don't want that here).
That's it! Also I got one of the limits of integration wrong. Thanks!
 
Top