# Thread: Area integral in polar co-ords of a surface

1. ## Area integral in polar co-ords of a surface

Express as an area integral in polar coordinates the area of the surface
z= y^2 +2xy -x^2 +2

lying over the annulus defined by 8/3 x^2 + y^2 1.
Hence evaluate this area in terms of π.

Thanks

2. Have you made any attempt at this at all?

The area you are to integrate over, in polar coordinates, is just $8/3\le r^2\le 1$ which, since r must be positive is exactly $\sqrt{8/3}\le r\le 1$. That makes the limits of integration easy. And don't forget that the "differential of area" in polar cordinates is $rdrd\theta$. Write $z= y^2+ 2xy- x^2+ 2$ in polar coordinates and integrate!

Ouch! I just realized this is a surface integral, not a volume integral. But the "polar coordinates" part is the same. To get the surface area integrand you can think of the surface $z= y^2+ 2xy- x^2+ 2$ as given by the vector function $\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (y^2- 2xy- x^2+ 2)\vec{k}$ with the x and y coordinates as parameters. The derivative vectors $\vec{r}_x= \vec{i}- (2y+ 2x)\vec{k}$ and $\vec{r}_y= \vec{j}+ (2y- 2x)\vec{k}$ are tangent to the surface and their cross product,
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -(2y+2x) \\ 0 & 1 & 2y- 2x\end{array}\right|= -(2y+ 2x)\vec{i}- (2y+ 2x)\vec{j}+ \vec{k}$
is perpendicular to the surface and gives the "vector differential of surface area". Its length, $\sqrt{1+ 8x^2+ 8y^2}$ gives the "scalar differential of surface area", $\sqrt{1+ 8x^2+ 8y^2}dx dy$. I think you can see that will be much easier to integrate in polar coordinates!