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

Answer :

Geophilip

Answer:

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

azikennamdi

The mathematical theorem which is relevant to this problem is called the Variable Change theorem or formula.

What is the Variable Change Form theorem?

The Variable Change Theorem is a theorem that explains how lengths, areas, volumes, and generalized n-dimensional volumes are skewed by functions that can be differentiated.

The step-by-step solution is given as follows

Information given about the boundary curve map is such that:

V = 0 = T(u, 0) = (U², 0), the line y = 0

V = U == T(u, u) = (0, 2U²), the line x = 0

Recall that:

U = 2 = T(2, v) = (4 - V², 4V).

If  x = 4 - V² and y = 4V, V = Y/4

Eliminating t will become:

X = 4 - (y/4)² → y² = 64 - 16x.

It, therefore, derives that T transforms the triangle to D.

Because x = U² - V² and Y = 2UV, we can say

∂(x,y)/∂(u,v) = |2u -2v|

|2v 2u| = 4(U² + V²).

Hence the

∫∫D √(x² + y²) dx dy

= ∫(u = 0 to 4) ∫(v = 0 to u) √[(u² - v²)² + (2uv)²] * 4(u² + v²) dv du

= ∫(u = 0 to 2) ∫(v = 0 to u) √(u² + v²)² * 4(u² + v²) dv du

= ∫(u = 0 to 2) ∫(v = 0 to u) 4(u² + v²)² dv du

= ∫(u = 0 to 2) ∫(v = 0 to u) 4(u⁴ + 2u² v² + v⁴) dv du

= ∫(u = 0 to 2) 4(u⁴ v + 2u⁴ v³/3 + v⁵/5) {for v = 0 to u} du

=4((u⁵ + 2u⁵/3 + u⁵/5) - (0))

= ∫(u = 0 to 4) (112/15) u⁵) du

= (112/15) u⁶/6 {for u = 0 to 2}

= 112 * ((2⁶)/90)

= 112 * 0.71111111111

= 79.6444444443

≈ 79.6

Learn more about Variable Change Theorem at

https://brainly.com/question/24563834

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