# Thread: surface area

1. ## surface area

Find surface area:
$z^2=2xy$
$x+y\leq1,x\geq0,y\geq0$

Got such integral
$\int\limits_0^1dx\int\limits_{1-x}^0 \sqrt{\frac{x^2+y^2+xy}{xy}}dy$

Any suggestions to solve it?

2. Originally Posted by waytogo
Find surface area:
$z^2=2xy$
$x+y\leq1,x\geq0,y\geq0$

Got such integral
$\int\limits_0^1dx\int\limits_{1-x}^0 \sqrt{\frac{x^2+y^2+xy}{xy}}dy$

Any suggestions to solve it?
$V=\iint\limits_T z\ dy\ dx$

$=\int\limits_0^1\int\limits_0^{1-x}\sqrt{2xy}\ dy\ dx$

3. Originally Posted by alexmahone
$V=\iint\limits_T z\ dy\ dx$

$=\int\limits_0^1\int\limits_0^{1-x}\sqrt{2xy}\ dy\ dx$
I think formula you have written is for volume, but I need a surface area. Formula for that in Cartesian coordinates is $S=\iint\limits_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}} dx dy$. That is from where my definite integral comes.

4. Originally Posted by waytogo
I think formula you have written is for volume, but I need a surface area. Formula for that in Cartesian coordinates is $S=\iint\limits_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}} dx dy$. That is from where my definite integral comes.
I'm sorry; I misread the question.

5. Can you tell us how you obtained your integrand $\sqrt{\frac{x^2+y^2+xy}{xy}}$?

6. Thank you, Danny! I rechecked my calculus and found a mistakte. So the actual integrand is $\frac{1}{\sqrt{2}}\iint\limits_D\frac{x+y}{\sqrt{x y}}dxdy=\frac{1}{\sqrt{2}}\int\limits_0^1dx\int\li mits_0^{1-x}\frac{x+y}{\sqrt{xy}}dy=...=\frac{2}{\sqrt{2}}\i nt\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx$. And this is where I have stopped for now, because I cannot deal with this integrand either! Any ideas?

7. Originally Posted by waytogo
Thank you, Danny! I rechecked my calculus and found a mistakte. So the actual integrand is $\frac{1}{\sqrt{2}}\iint\limits_D\frac{x+y}{\sqrt{x y}}dxdy=\frac{1}{\sqrt{2}}\int\limits_0^1dx\int\li mits_0^{1-x}\frac{x+y}{\sqrt{xy}}dy=...=\frac{2}{\sqrt{2}}\i nt\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx$. And this is where I have stopped for now, because I cannot deal with this integrand either! Any ideas?
$\int\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx$

Let $x=u^2 \to 1-x=1-u^2 \to dx=2udu$

$\frac{2}{3} \int (2u^2+1) \sqrt{1-u^2} du = \frac{4}{3} \int u^2 \sqrt{1-u^2}du + \frac{2}{3} \int \sqrt{1-u^2}du$

You should be able to evaluate both of these easily (sin substitution should work).

8. yeah, it always seems easy, when someone gives a clue. thanks, AllanCuz!