# Thread: Area Integral

1. ## Area Integral

I have the following question:

Express as an area integral in polar co-ordinates the area of the surface z = 5 + xy lying over the annulus defined by 2 < x^2 + y^2 < 3. Hence evaluate the area in terms of pi.

I know i have to get the function:

g(r,0) = sqrt 1 + (df/dr)^2 + (df/d0)^2

then evaluate it over the interval 2 to 3.

The problem is i cant seem to get the required answer 13.628pi. All my answers are nuts!!!

are the partial derivatives 0^2 and r^2 ?

and does the 5 in the function get added to the 1 in g(x,y)?

In other words should I have

g(r, 0) = sqrt 6 + 0^2 + r^2 ??

N.B. 0 is meant to be theta.

2. Originally Posted by iwishiunderstood
I know i have to get the function:

g(r,0) = sqrt 1 + (df/dr)^2 + (df/d0)^2

then evaluate it over the interval 2 to 3.
Are you sure you do? Your formula is rather more reminiscent of a computation of arc length, or kind of.
To compute the area of a surface parametrized by $(u,v)\mapsto M(u,v)$, you need to integrate the "surface element" $\left\|\frac{\partial M}{\partial u}\times\frac{\partial M}{\partial v}\right\|du\, dv$ (where $\times$ is the cross product of vectors) over the range of $u$ and $v$ (this is a double integral). In your case, $r$ and $\theta$ replace $u$ and $v$, and $M(r,\theta)=(r\cos\theta,r\sin\theta,5+(r\cos\thet a)(r\sin\theta))$ where $r\in[\sqrt{2},\sqrt{3}]$ and $\theta\in[0,2\pi]$. I haven't tried much to compute the integral I obtained this way, but it doesn't look very nice anyway...

I hope you'll be able to carry on using this.

Laurent.

(I forgot to mention that a more convenient expression for the surface element is $\sqrt{EG-F^2} du\,dv$ where $E=\left\|\frac{\partial M}{\partial u}\right\|$, $G=\left\|\frac{\partial M}{\partial v}\right\|$ and $F=< \frac{\partial M}{\partial u},\frac{\partial M}{\partial v}>$)

3. definitely sure - I have the notes for the assignment from my tutor and these say(as well as the text book) to find g(x,y) in polar co-ordinates.

4. OK. In fact, my formula reduces, in polar coordinates, to: $\int_0^{2\pi}\int_{\sqrt{2}}^{\sqrt{3} }g(r,\theta)dr\,d\theta$, where $g(r,\theta)=\sqrt{r^2+r^2 (\partial_r f)^2+(\partial_\theta f) ^2}$. This formula with the $r^2$ is homogeneous to an area, while yours is not ( $f$ is a length, so that $\partial_r f$ is homogeneous to a constant and $\partial_\theta f$ is homogeneous to a length).

And in the present case $f(r,\theta)=5+xy=5+r^2\cos\theta\sin\theta$.

This gives: $g(r,\theta)=\sqrt{r^2+r^4}$, so the area is $2\pi\int_{\sqrt{2}}^{\sqrt{3}}\sqrt{r^2+r^4}dr=2\p i\int r\sqrt{1+r^2}dr$ $=\frac{2}{3}\pi [(1+r^2)^{3/2}]_{\sqrt{2}}^{\sqrt{3}}=\frac{2}{3}\pi((1+3)^{3/2}-(1+2)^{3/2})=\frac{2}{3}(8-\sqrt{27})\pi$. There must be a mistake, but the idea is there.

5. Originally Posted by Laurent
There must be a mistake, but the idea is there.
The mistake is in the statement of the problem: the annulus must be given by $2^2\leq x^2+y^2\leq 3^2$, so that the integration is between $2$ and $3$, and you get that the area is $\pi$ times $\frac{2}{3}(10^{3/2}-5^{3/2})=\frac{2}{3}(\sqrt{1000}-\sqrt{125})=\frac{10\sqrt{5}}{3}(2\sqrt{2}-1)\simeq 13.628$. This way you have the expected result.

6. thanks for that - quality bit of help. cheers!!!

can i just ask how you got the srt r^2 + r^4 in the answer please?