use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (2, 1), and

Question

1 year ago

6

(1, 2). $ l

1 answer:

Maksim231197 [3] · Answer 1 · 2023-09-15T11:40:37+03:00

After evaluation, The value of the integral is -3.

Calculation:

The triangular region R with vertices (0,0), (2, 1) and (1,2)

And also the transformation is x = 2u+v, y= u+2v

The sketch of the region with vertices A(0,0), B(2,1), and C(1, 2) is as follows:

The straight line passing through P(x₁,y₁) and Q(x₂,y₂) is

$\frac{x-x_{1 } } {x_{2} -x_{1} } =\frac{y-y_{1 } } {y_{2} -y_{1} }$

The line R₁ passes through points A(0,0) and B(2,1)

So, the equation of the line R₁ is

$\frac{x-0 } {2-0} =\frac{y-0 } {1 -0 }$

⇒ $y=\frac{x}{2}$

The line R₂ passes through points B(2,1) and C(1,2).

So, the equation of the line R₂ is

$\frac{x-2 } {1 -2 } =\frac{y-1 } {2 -1 }$

y=3-x

The line R₃ passes through the points C(1,2) and A(0,0): So, the equation of the line R₃ is

$\frac{x-1 } {0 -1 } =\frac{y-2 } {0 -2 }$

y=2x

The transformation of R₁ is x=2u+v,y=u+2v

$y=\frac{x}{2}$

$u+2v=\frac{1}{2} (2u+v)$

v=0 ------- (1)

The transformation on R₂ is x=2u+v,y=u+2v

x+y=3

(2u+v)+(u+2v)=3

u+v=1 ---------(2)

The transformation on R₃ is x=2u+v,y=u+2v

y=2x

u+2v=2(2u+v)

u=0 -------(3)

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