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Homework Help: Integral: Area of a surface in 3D

  1. Dec 18, 2005 #1
    The problem goes like that: we have a curve, z = xy, and we want to find the portion of the surface area above the circle (x^2 + y^2 <= 1 in the (x,y)-plane).

    It's probably involving a double integral in polar coordinate or a triple in cylindrical, but I don't know how to set the problem. I did some volume under surface in 3D, but I don't know how to calculate the area of this surface.

    Thank you very much!
  2. jcsd
  3. Dec 19, 2005 #2
    Does this look familliar?

    (summarised from Thomas' Calculus, Section 13.6)

    In general, if you have a surface [itex]S[/itex] which is paramaterized in terms of two variables [itex]u,v[/itex] such that you can represent the surface with a function [itex]\vec{r}(u,v) = f(u,v) \hat{i} + g(u,v) \hat{j} + h(u,v) \hat{k}[/itex], then you can calculate its surface area as:

    [tex]\int_a^b \int_c^d \left| \vec{r_u} \times \vec{r_v} \right| du \, dv[/tex]

    where [itex]\vec{r_u}[/itex] and [itex]\vec{r_v}[/itex] are the partial derivatives of [itex]\vec{r}(u,v)[/itex] with respect to [itex]u[/itex] and [itex]v[/itex], the vertical bars represent the magnitude of the vector product, and [itex]a,b,c,d[/itex] are the limiting values for [itex]u,v[/itex].

    And you're right, it's easier if you convert to polar coordinates.
    Last edited: Dec 19, 2005
  4. Dec 19, 2005 #3


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    Homework Helper

    Surface Area

    The surface area of the portion of the surface [itex]z=f(x,y)[/itex] which lies above D is given by this integral:

    [tex]SA = \int\int_{D} \sqrt{1+ \left( \frac{\partial f}{\partial x} \right) ^{2} + \left( \frac{\partial f}{\partial y} \right) ^{2}} dA[/tex]

    Since D denotes the region in the xy-plane given by:

    [tex]D=\left\{ (x,y)\in\mathbb{R} ^2 : x^2+y^2\leq 1\right\}[/tex]

    i.e., the unit disk. And we have [itex]z=xy[/itex], it follows that

    [tex]SA = \int\int_{D} \sqrt{1+ \left( \frac{\partial f}{\partial x} \right) ^{2} + \left( \frac{\partial f}{\partial y} \right) ^{2}} dA = \int\int_{x^2+y^2\leq 1} \sqrt{1+ y ^{2} + x^{2}} dxdy [/tex]

    Change to polar coordinates to get...

    [tex]SA = \int_{\theta=0}^{2\pi} \int_{r=0}^{1} \sqrt{1+ r^{2}} r drd\theta = \int_{\theta=0}^{2\pi}d\theta \int_{u=1}^{2} \sqrt{u} \frac{du}{2} = \frac{2\pi}{3} \left( 2\sqrt{2} - 1 \right) [/tex]
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