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Solve the system $$u = 3x + 2y, v = x + 4y$$ to find expressions for $x$ and $y$ in terms of $u$ and $v$.

Use these expressions to find the Jacobian $∂(x, y)/∂(u, v)$.

Hence evaluate the integral $$\iint(3x + 2y)(x + 4y) dx dy$$ for the region $R$ bounded by the lines $$y = −(3/2)x + 1,\ y = −(3/2)x + 3$$ and $$y = −(1/4)x,\ y = −(1/4)x + 1$$

So I computed the Jacobian = 1/10, I solved the equations to find $x = h(u,v)$ and $y = g(u,v)$ and then I substituted $h$ and $g$ in the equations for the boundaries ,which gave me $u = 2$ and $u = 6$, and $v = 0$ and $v = 4$. So the integral I have to compute is equal to $$\iint uvJ(u,v)dudv$$ with the above boundaries?

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  • $\begingroup$ My comment was the one that was mistaken, sorry. Your notes are probably right. $\endgroup$
    – evaristegd
    Mar 24, 2019 at 1:55
  • $\begingroup$ So is the integral I have to compute equal to $∬uvJ(u,v)dudv$, with the boundaries I found? $\endgroup$
    – user600210
    Mar 24, 2019 at 1:58

1 Answer 1

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Yes, it is correct. The answer is $12.8$.

Direct calculation:

$\hspace{1cm}$enter image description here

$$\int_{0}^{0.8}\int_{-\frac32x+1}^{-\frac14x+1} (3x+2y)(x+4y) dydx+\\ \int_{0.8}^{1.6}\int_{-\frac14x}^{-\frac14x+1} (3x+2y)(x+4y) dydx+\\ \int_{1.6}^{2.4}\int_{-\frac14x}^{-\frac32x+3} (3x+2y)(x+4y) dydx=\\ \frac{44}{15}+\frac{104}{15}+\frac{44}{15}=\frac{192}{15}=12.8.$$

Your method: $$\int_{0}^{4}\int_{2}^{6} \frac1{10}uv dudv=\int_0^4 \frac85vdv=12.8.$$

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  • $\begingroup$ I got the same result, thanks for the work! $\endgroup$
    – user600210
    Mar 25, 2019 at 22:19
  • $\begingroup$ You are welcome. Good luck. $\endgroup$
    – farruhota
    Mar 26, 2019 at 2:11

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