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
Let $\displaystyle A$ be your area, then
$\displaystyle A=\iint_Rds,$ where $\displaystyle R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\}$ and $\displaystyle ds = \sqrt {1 + \left( {\frac{{\partial z}}
{{\partial x}}} \right)^2 + \left( {\frac{{\partial z}}
{{\partial y}}} \right)^2 } \,dy\,dx.$
Can you proceed from there?
Careful when writting the integration order.
I wrote it in the order $\displaystyle dy\,dx$ so that you can evaluate directly the double integral. Since you wrote it in the order $\displaystyle dx\,dy$ you have to reverse integration order, so we require $\displaystyle 0\le y\le1,\,\sqrt[3]y\le x\le1.$
Evaluating the double integral with $\displaystyle dx\,dy$ or $\displaystyle dy\,dx$ order, both answers are in agreement.