# Thread: [HELP!] Parametric Surface Area

1. ## [SOLVED] Parametric Surface Area

I'm so stuck with the Parametric Surface Area problem right now T_T

There are two questions in total, if you can help me with any one of the two, it'll be so appreciated!

Question 1:
The part of the plane 3x+2y+z = 6 that lies in the 1st octant.

Question 2:
The surface z = 2/3 (x^(3/2) + y^(3/2)), 0 <= x <= 1, 0 <= y <= 1

2. For reference, the formula for surface area is

$\int\int_R \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dA.$

3. Originally Posted by Scott H
For reference, the formula for surface area is

$\int\int_R \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dA.$
I know this equation, but what are the boundaries? and do I integrate with respect to x and y or r and (theta)?

4. Because the region is defined by

$0\le x \le 1$
$0\le y \le 1,$

the integral for surface area will be

$\int_0^1\int_0^1 \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dx\,dy.$

As we are given a square region and a simple integrand (when you simplify it, that is!), we can make do with Cartesian coordinates in this integral.

5. Originally Posted by Scott H
Because the region is defined by

$0\le x \le 1$
$0\le y \le 1,$

the integral for surface area will be

$\int_0^1\int_0^1 \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dx\,dy.$

As we are given a square region and a simple integrand (when you simplify it, that is!), we can make do with Cartesian coordinates in this integral.
OHHH~~get it now~!
thank you so much!! =D

6. Not a problem.

7. Originally Posted by violet8804
I'm so stuck with the Parametric Surface Area problem right now T_T