Show that T(u, v)=(u2−v2,2uv)T(u, v)=(u2−v2,2uv) maps the triangle ={(u, v):0≤v≤u≤2}={(u, v):0≤v≤u≤2} to the domain DD bounded by x=0x=0, y=0y=0, and y2=64−16xy2=64−16x. Use TT to evaluate

79.64

Step-by-step explanation:

Consider the boundary curve map

Since

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

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

Since

u = 2 == T(2, v) = (4 - v^2, 4v).

Since x = 4 - v^2 and y = 4v,

V=y/4

eliminating t yields x = 4 - (y/4)^2 ==> y^2 = 64 - 16x.

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 2) ∫(v = 0 to u) √(u^2 + v^2)^2 * 4(u^2 + v^2) dv du

= ∫(u = 0 to 2) ∫(v = 0 to u) 4(u^2 + v^2)^2 dv du

= ∫(u = 0 to 2) ∫(v = 0 to u) 4(u^4 + 2u^2 v^2 + v^4) dv du

= ∫(u = 0 to 2) 4(u^4 v + 2u^2 v^3/3 + v^5/5) {for v = 0 to u} du

=4((u^5 + 2u^5/3 + u^5/5) - (0))

= ∫(u = 0 to 4) (112/15) u^5) du

= (112/15) u^6/6 {for u = 0 to 2}

= 112 * 2^6/90

= 7166/90 = 358/45 = 79.64

the answer to this question is yes

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step-by-step explanation: