Double integral help?

The region R is enclosed by the x axis and the three parabolas


x = (y^2/4) - 1


x = 1 - (y^2/4)


x = 4 - (y^2/16)





Note that R is the image of the rectangle a ≤ u ≤ b, 0 ≤ v ≤ c in the uv-plane under the map f = %26lt;x, y%26gt; = %26lt;u^2 - v^2, 2uv%26gt;.





Use f to do a coordinate change that will let you compute the following double integral.





∫∫ x dxdy


R

Double integral help?
Interesting. I would have done this by horizontal cross-sections:





Int {0 : y : 2, 1-y^2/4 : x : 4-y^2/16} x dx dy +


Int {2 : y : 4, y^2/4 - 1 : x : 4-y^2/16} x dx dy





The other approach you've suggested may provide a more elegant solution, but it's not familiar to me. So far, all I have is a=1, b=2, c=1. Here's my plan:


1) Take a small square (du by dv) in the uv-plane;


2) Run it through the transformation to find its xy-image;


3) Compute the area of that image, in terms of u and v;


4) Multiply by x=u^2-v^2;


5) Integrate.





Let's move up to step 3. The image of the small square is a nearly rectangular thing; we'd like to know how long its sides are.





Start with equations


x = u^2 - v^2, y = 2uv.


Take u and v partials:


x_u = 2u


y_u = 2v


These two give us the length of one side of the rectangle: Sqrt(4u^2 + 4v^2)





x_v = -2v


y_v = 2u


These two give us the length of the other side of the rectangle: Sqrt(4v^2 + 4u^2).





Happily, these are the same; thus, the area of the almost-rectangle is their product, 4u^2 + 4v^2. So,





dA = dx dy = 4(u^2+v^2) du dv.





We now multiply this by x = u^2-v^2 to get





∫∫ x dxdy = ∫∫ 4(u^2 + v^2)(u^2 - v^2) du dv


R





= 4 ∫∫ u^4 - v^4 du dv


= 4 ∫ [u^5 / 5 - uv^4]_1^2 dv


= 4 ∫ [6.2 - v^4] dv


= 4 [6.2 v - v^5 / 5]_0^1


= 4 [6.2 - .2]


= 24





Whew.





Edit: while checking this answer by way of my horizontal cross-section idea, I realized something.





Let region A be the one in question (in the first quadrant).


Let region B be the region bounded by x = 1- y^2/4 and x = y^2/4 - 1, for 2 %26lt; y %26lt; 4.


Let region C be the region bounded by y=0, y=4, x = 1 - y^2/4, x=4 - y^2/16.





Then (integral over A) + (integral over B) = (integral over C).


But region B is symmetric with respect to the y-axis, so (integral over B) = 0.





The integral over C has only one piece, which is nice. That integral works out to 24 as well, so at this point I'm pretty confident in my answer.