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

16.9 times 2.85

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b