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...

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