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Kazeer [188]
3 years ago
5

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

(1, 8). (x − 3y) da r , x = 8u + v, y = u + 8v
Mathematics
1 answer:
Jlenok [28]3 years ago
6 0
We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

\frac{y}{x} = \frac{1}{8}  \\  \\ \Rightarrow x=8y \\  \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\  \\ \Rightarrow v=0

For the line with points (0, 0) and (1, 8), the equation is given by:

\frac{y}{x} = \frac{8}{1}  \\  \\ \Rightarrow y=8x \\  \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\  \\ \Rightarrow u=0

For the line with points (8, 1) and (1, 8), the equation is given by:

\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\  \\ \Rightarrow y-1=-x+8 \\  \\ \Rightarrow y=-x+9 \\  \\ \Rightarrow u+8v=-8u-v+9 \\  \\ \Rightarrow u=1-v

The Jacobian determinant is given by

\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\  \\ =64-1=63

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\  \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\  \\ =63(-9)=|-576|=576
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