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Math Help - [HELP!] Parametric Surface Area

  1. #1
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    Exclamation [SOLVED] Parametric Surface Area

    I'm so stuck with the Parametric Surface Area problem right now T_T
    Can anyone please help me out?

    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

    Please, help!
    Last edited by violet8804; December 10th 2009 at 08:30 AM. Reason: Question is solved.
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    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)?
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by Scott H View Post
    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
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  6. #6
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    Not a problem.
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  7. #7
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    Quote Originally Posted by violet8804 View Post
    I'm so stuck with the Parametric Surface Area problem right now T_T
    Can anyone please help me out?

    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.

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

    Please, help!
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