1. ## surface integral

Hello, I need help with this one.

Calculate the area of the surface defined by:
z=sqrt(2xy), where 0<= x <=1 and 0<= y <= x^3

/Alex

2. Let $A$ be your area, then

$A=\iint_Rds,$ where $R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\}$ and $ds = \sqrt {1 + \left( {\frac{{\partial z}}
{{\partial x}}} \right)^2 + \left( {\frac{{\partial z}}
{{\partial y}}} \right)^2 } \,dy\,dx.$

Can you proceed from there?

3. Originally Posted by Krizalid
Let $A$ be your area, then

$A=\iint_Rds,$ where $R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\}$ and $ds = \sqrt {1 + \left( {\frac{{\partial z}}
{{\partial x}}} \right)^2 + \left( {\frac{{\partial z}}
{{\partial y}}} \right)^2 } \,dy\,dx.$

Can you proceed from there?
I've got this far.
$
Zx' = \frac{y}{\sqrt{2xy}}$

$
Zy' = \frac{x}{\sqrt{2xy}}

$

=>

$
A = \iint_R \sqrt{1 + (\frac{y}{\sqrt{2xy}})^2 +(\frac{x}{\sqrt{2xy}})^2} dxdy = \iint_R \sqrt{\frac{x^2}{2xy} + \frac{y^2}{2xy} + 1} dxdy$

4. Originally Posted by yOdA
I've got this far.
$
Zx' = \frac{y}{\sqrt{2xy}}$

$
Zy' = \frac{x}{\sqrt{2xy}}

$

=>

$
A = \iint_R \sqrt{1 + (\frac{y}{\sqrt{2xy}})^2 +(\frac{x}{\sqrt{2xy}})^2} dxdy = \iint_R \sqrt{\frac{x^2}{2xy} + \frac{y^2}{2xy} + 1} dxdy$
$\sqrt{\frac {x^2}{2xy} + \frac {y^2}{2xy} + 1} = \sqrt{\frac {x^2 + y^2 + 2xy}{2xy}} = \sqrt{\frac {(x + y)^2}{2xy}} = \frac {x + y}{\sqrt{2xy}} = \frac {\sqrt{x}}{\sqrt{2y}} + \frac {\sqrt{y}}{\sqrt{2x}}$

5. Careful when writting the integration order.

I wrote it in the order $dy\,dx$ so that you can evaluate directly the double integral. Since you wrote it in the order $dx\,dy$ you have to reverse integration order, so we require $0\le y\le1,\,\sqrt[3]y\le x\le1.$

Evaluating the double integral with $dx\,dy$ or $dy\,dx$ order, both answers are in agreement.