Evaluate the double integrable by making the appropriate change of variables

Question

a) Evaluate $\iint_D \frac{1}{y}~dA$, where D is the region bounded by $y^3=x^2$, $y^3=6x^2$, $y=2x$, $y=3x$ in the first quadrant By changing the variables I get $u=\frac{y^3}{x^2}$ and $v=\frac{y}{x}$. Then for the Jacobian I get J= $\frac{1}{v^5}$. Through Change of Variables I get $\iint_D \frac{1}{y}~dA$= $$\int_{2}^{3} \int_{1}^{6}\frac{1}{v^5}\frac{v^2}{u}dudv=\ \int_{2}^{3} \int_{1}^{6} \frac{1}{uv^3}dudv$$ Evaluted I get $ \frac{5(ln 6 - ln 1)}{72}$. Is this correct?

Answer 1

Are you sure Jacobian is right?

after changing variables to

$u = y^3 / x^2$ and $v = y/x$

we get $y = u/v^2$ and $x = u/v^3$

now the derivatives are

$$\frac{\partial x}{\partial u} = {1\over v^3},\quad\frac{\partial y}{\partial u} = {1\over v^2}$$ $$\frac{\partial x}{\partial v}=-{3u\over v^4},\quad\frac{\partial y}{\partial v} = -{2u\over v^3}$$

and so Jacobian is $$J = -{2u\over v^6} + {3u\over v^6} = {u\over v^6}$$

now the integration

$$∫_2^3dv∫_1^6{v^2\over{u}}{u\over{v^6}}du= ∫_2^3dv∫_1^6{1\over v^4}du=5∫_2^3{1\over v^4}dv = 5(-{1\over 5v^5})_2^3 = {1\over 2^5}-{1\over 3^5} $$