Integral: Area of a surface in 3D

[LeonL]

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!

 

[dicerandom]

Does this look familliar?

(summarised from Thomas' Calculus, Section 13.6)

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

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

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

And you're right, it's easier if you convert to polar coordinates.

 

[benorin]

Surface Area

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

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

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

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

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

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

Change to polar coordinates to get...

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)