# [HELP!] Parametric Surface Area

• Dec 9th 2009, 01:54 PM
violet8804
[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

• Dec 9th 2009, 02:21 PM
Scott H
For reference, the formula for surface area is

$\displaystyle \int\int_R \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dA.$
• Dec 9th 2009, 03:12 PM
violet8804
Quote:

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

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

I know this equation, but what are the boundaries? and do I integrate with respect to x and y or r and (theta)?
• Dec 9th 2009, 03:58 PM
Scott H
Because the region is defined by

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

the integral for surface area will be

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

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.
• Dec 9th 2009, 04:07 PM
violet8804
Quote:

Originally Posted by Scott H
Because the region is defined by

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

the integral for surface area will be

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

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
• Dec 9th 2009, 04:16 PM
Scott H
Not a problem. :)
• Dec 10th 2009, 05:01 AM
HallsofIvy
Quote:

Originally Posted by violet8804
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.

The plane 3x+ 2y+ z= 6 crosses the xy-plane, where z= 0, in the line 3x+ 2y= 6. That, in turn, has y-intercept (0, 3) and x-intercept (2, 0). You can take the limits of integration to be x=0 to 2 and, for each x, y= 0 to 3- (3/2)x. Or, if you prefer to integrate with respect to x first, y from 0 to 3 and, for each y, x from 0 to 2- (2/3)y.

Quote:

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