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I need help in understanding how to compute the cross product of two partial derivatives to help me calculate a surface area. I've watched the Khan Academy lecture on the subject but they seem to be about functions resulting in vectors, while I'm working with functions resulting in scalars (if I'm talking nonsense, I am sorry).

I am given the function

$$f(x,y) = \sqrt{x^2 + y^2}$$

and I've computed its partial derivatives to be

$$\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}$$

$$\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}$$

How do I calculate the cross product $\frac{\partial f}{\partial x} \times \frac{\partial f}{\partial y}$?

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1 Answer 1

up vote 3 down vote accepted

I'll take a guess about what you're trying to do.

As you write, you've got a scalar function of two variables. Since you say you're trying to calculate a surface area, we have to somehow interpret this as defining a surface. A straightforward way to do that would be to take this function as the $z$ component of a surface over the $x$-$y$ plane.

We can treat this surface just like any parameterized surface by regarding $x$ and $y$ as parameters which happen to coincide with two of the coordinates. Then the parametrization of the surface is

$$\vec r (x,y) = (x,y,f(x,y))=\left(x,y,\sqrt{x^2+y^2}\right)\;.$$

Then the tangent vectors along the coordinate lines are

$$\frac{\partial\vec r(x,y)}{\partial x}=\left(1,0,\frac x{\sqrt{x^2+y^2}}\right)$$

and

$$\frac{\partial\vec r(x,y)}{\partial y}=\left(0,1,\frac y{\sqrt{x^2+y^2}}\right)\;,$$

and the surface element is

$$ \begin{eqnarray} \mathrm dS &=& \left|\frac{\partial\vec r(x,y)}{\partial x}\times\frac{\partial\vec r(x,y)}{\partial y}\right|\,\mathrm dx \mathrm dy \\ &=& \left|\left(-\frac x{\sqrt{x^2+y^2}},-\frac y{\sqrt{x^2+y^2}},1\right)\right|\,\mathrm dx \mathrm dy \\ &=& \sqrt2\mathrm dx \mathrm dy\;. \end{eqnarray} $$

What this tells you is that the surface element of a cone at an angle $45^\circ$ (which is the surface described by that function) is $\sqrt2$ (the reciprocal of the sine of the cone's angle) times the projected surface element in the $x$-$y$ plane.

I hope that's relevant to what you had in mind.

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Thank you! Makes perfect sense. :) –  Paul Manta Jan 22 '12 at 3:03

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