Please help!! Change of variables?

2012-03-09 1:22 am
http://img140.imageshack.us/img140/5613/58847494.jpg
Show that T(u,v) = (u^2 - v^2, 2uv) maps the triangle = {(u,v): 0 ≤ v ≤ u ≤ 4} to the domain D bounded by x=0, y=0, and y^2 = 1024 - 64x.

Use T to evaluate ∬D sqrt(x^2+y^2) dxdy

回答 (1)

2012-03-09 5:22 am
✔ 最佳答案
First, check that boundary curves map to boundary curves.

v = 0 ==> T(u, 0) = (u^2, 0), effectively the line y = 0

v = u ==> T(u, u) = (0, 2u^2), effectively the line x = 0

u = 4 ==> T(4, v) = (16 - v^2, 8v).
Since x = 16 - v^2 and y = 8v, eliminating t yields x = 16 - (y/8)^2 ==> y^2 = 1024 - 64x.

Hence T transforms the triangle to D.
---------------
Since x = u^2 - v^2 and y = 2uv,
∂(x,y)/∂(u,v) =
|2u -2v|
|2v 2u| = 4(u^2 + v^2).

Therefore, change of variables yields
∫∫D √(x^2 + y^2) dx dy
= ∫(u = 0 to 4) ∫(v = 0 to u) √[(u^2 - v^2)^2 + (2uv)^2] * 4(u^2 + v^2) dv du
= ∫(u = 0 to 4) ∫(v = 0 to u) √(u^2 + v^2)^2 * 4(u^2 + v^2) dv du
= ∫(u = 0 to 4) ∫(v = 0 to u) 4(u^2 + v^2)^2 dv du
= ∫(u = 0 to 4) ∫(v = 0 to u) 4(u^4 + 2u^2 v^2 + v^4) dv du
= ∫(u = 0 to 4) 4(u^4 v + 2u^2 v^3/3 + v^5/5) {for v = 0 to u} du
= ∫(u = 0 to 4) (112/15) u^5) du
= (56/45) u^6 {for u = 0 to 4}
= 229376 / 45.

I hope this helps!


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