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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). enter image description here

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

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

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  • $\begingroup$ Would you advice me to convert it into polar co-ordinates or integrate it as you've set up above? $\endgroup$ – Prathik Gurudatt Jun 24 '17 at 1:29

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