Question on a double integral with change of variables

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?

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

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

Direct calculation:

$$\hspace{1cm}$$

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

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

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