The prompt here is to find the surface area using double integrals.
$$f(x) = 2\sqrt{xy}$$
with the vertices (1,1) (1,2) (2,2) (2,1).
From resources the formula for surface area using double integrals is $$ A = \int_1^2 \int_1^2\sqrt{1 + (\partial/\partial x)^2 + (\partial/\partial y )^2}\,dx \, dy $$
Now the question is, when we find the partial derivative of the each of the terms inside for the above function,
$$\partial / \partial x = 2 \sqrt x \sqrt y$$ can this be split up like this or $$\partial / \partial x = 2 \sqrt {x y}$$
are there any other methods to solve the integral with the square root? One of my previous post had a similar question but with volume to be found, one user had split the root in the above manner. But on use they give different results.
What could be a better option or method to solve the integral.
Update: I tried solving it like this. $$\int_1^2 \int_1^2 ( 1 + (x^2 + y^2)/xy)rdrd\theta$$ $$\int_1^2 \int_1^2 ( 1 + (r^2)/xy)rdrd\theta$$ $$\int_1^2 \int_1^2 ( 1 + (\frac {r^2} {rcos\theta rsin\theta}) rdrd\theta$$ $$ \int_1^2 \int_1^2 (1 + (\frac {r} {cos\theta \ sin\theta}) rdrd\theta $$
In teh given domain we have $x>0$ and $y>0$, so $$ 2\sqrt{x}\sqrt{y}=2\sqrt{xy} $$ and the derivatives are: $$ \frac{\partial f}{\partial x}=\sqrt{\frac{y}{x}} \qquad \frac{\partial f}{\partial y}=\sqrt{\frac{x}{y}} $$
but the I suspect that the integral $$ \int_1^2 \int_1^2\sqrt{1+\frac{y}{x}+\frac{x}{y}}dx dy $$
cannot be calculated with elementary functions.