By a change of variable x(u, v) = uv, y(u, v) = v/u is double integral, the integrand f(x, y) changes to f(uv, v/u) \[\phi \] (u, v). Then, \[\phi \] (u, v) is A. 2 u/v B. 2 uv C. v2 D. 1

2 u/v
2 uv
v2
1

The correct answer is $\phi(u, v) = 1$.

To see this, let us consider a double integral over the region $R$ in the $xy$-plane:

$$\iint_R f(x, y) \, dxdy$$

We can write this integral in terms of the new variables $u$ and $v$ as follows:

$$\iint_R f(uv, v/u) \phi(u, v) \, dudv$$

where $\phi(u, v)$ is a Jacobian of the transformation from the $xy$-plane to the $uv$-plane.

The Jacobian is a determinant

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that tells us how much the area of a small rectangle in the $xy$-plane changes when we transform it to the $uv$-plane. In this case, the Jacobian is equal to $1$, so we can write the integral as follows:

$$\iint_R f(uv, v/u) \, dudv = \iint_R f(x, y) \, dxdy$$

Therefore, $\phi(u, v) = 1$.