# surface area using formula and double integrals

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. $$\int_1^2 \int_1^2 ( 1 + (x^2 + y^2)/xy)rdrd\theta$$ $$\int_1^2 \int_1^2 ( 1 + (r^2)/xy)rdrd\theta$$ $$\int_1^2 \int_1^2 ( 1 + (\frac {r^2} {rcos\theta rsin\theta}) rdrd\theta$$ $$\int_1^2 \int_1^2 (1 + (\frac {r} {cos\theta \ sin\theta}) rdrd\theta$$

• $\frac {\partial f}{\partial x} =\sqrt{\frac yx}.$ What do you suppose$\frac {\partial f}{\partial y}$ is? – Doug M Jun 23 '17 at 15:44
• The partial derivative with respect to x. – Prathik Gurudatt Jun 23 '17 at 15:49

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 $$\int_1^2 \int_1^2\sqrt{1+\frac{y}{x}+\frac{x}{y}}dx dy$$