Use the given transformation to evaluate the given integral, where R is the triangular region with vertices (0, 0), (6, 1), and (1, 6). (x - 3y) dA, X = 6u + v, y = u + 6v Step 1 For the transformation x = 6u + v, y =u + 6v, the Jacobian is ax 1 a(x, y) = du av = 35 %3D a(u, v) ду ду dv 1 ди Also, X - 3y = (6u + v) 3(u + 6v) = 3u - 17 - V. Step 2 To find the region S in the uv-plane which corresponds to R, we find the corresponding boundaries. The line though (0, 0) and (6, 1) is y = 1/6 x, and this is the image of v = 0 Step 3 The line through (6, 1) and (1, 6) is y = and this is the image of v = The line through (0, 0) and (1, 6) is y = and this is the image of u = 0.
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![**Use the given transformation to evaluate the given integral, where \( R \) is the triangular region with vertices (0, 0), (6, 1), and (1, 6).**
\[
\iint_R (x - 3y) \, dA, \quad x = 6u + v, \quad y = u + 6v
\]
**Step 1**
For the transformation \( x = 6u + v, \, y = u + 6v \), the Jacobian is
\[
\frac{\partial(x, y)}{\partial(u, v)} =
\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}
=
\begin{vmatrix}
6 & 1 \\
1 & 6
\end{vmatrix}
= 35
\]
Also,
\[
x - 3y = (6u + v) - 3(u + 6v) = 3u - 17v.
\]
**Step 2**
To find the region \( S \) in the \( uv \)-plane which corresponds to \( R \), we find the corresponding boundaries. The line through (0, 0) and (6, 1) is \( y = \frac{1}{6}x \), and this is the image of \( v = 0 \).
**Step 3**
The line through (6, 1) and (1, 6) is \( y = \) [blank], and this is the image of \( v = \) [blank].
The line through (0, 0) and (1, 6) is \( y = \) [blank], and this is the image of \( u = 0 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f8f1aa0-ff42-403f-89a3-b6ebbbc6b6ad%2F1105206f-6eea-422b-bbaa-beeda815a031%2Fmh98ap_processed.jpeg&w=3840&q=75)
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