Double integral (and for the area enclosed by a lemniscate).

Question

a) Transform into polar coordinates and compute the integral

$\int_\Omega ln(1+x^2+y^2)d(x,y)$

where $\Omega$ is the interior of the unit circle in the first quadrant.

b) $\Omega\subseteq R^2$ which is the subset enclosed by the $Lemniscate$ $(x^2+y^2)^2=4(x^2-y^2)$. Sketch $\Omega$ and compute the volume $v_2(\Omega)$.

Here is my approach to those:

a)First I transformed the general coordinates in the unit circle into polar coordinates. Since it's the unit circle I got $(rcos(\phi),rsin(\phi)),r=1$. I looked up the general formula for the integration in polar coordinates and got to $\int_\Omega ln(1+x^2+y^2)rdrd\phi$. Now, at this point I'm not so sure. Am I allowed to use the identity $cos^2(\phi)+sin^2(\phi)=1$ and plugging it into the argument of ln? Like $\int_\Omega ln(2)rdrd\phi$?

b) Now I am not sure how to approach this. I looked up what a Lemniscate is, it looks like the infinity-symbol centered around the origin. But I don't know how to get the volume of it. I suppose it's a double integral where $f(x,y)=1$, but I don't know how to set up the boundaries of the integrals and whether or not I should do it in cartesian or polar coordinates.

If someone had some tips or hints about these I would greatly appreciate it.

Are a and b related with each other? When it says "volume" does it mean surface integrate $ln(1+x^2+y^2)$? Is this the entirety of the question statement? — Gappy Hilmore, May 25 '15 at 08:44
Why is it called a "volume" and not a surface, do you have any idea? — Gappy Hilmore, May 25 '15 at 08:52
The prof introduced surfaces as 2-dimensional volumes. I guess that's why. — Konstantin, May 25 '15 at 08:54

score 4 · Accepted Answer · edited May 25 '15 at 12:59

$a)$ One thing you need to know by heart is this transformation: $x^2+y^2=r^2$, so the integrand becomes $\ln (1+r^2)$. We also have $dx\,dy=rdr\,dt$ (I will use $t$ instead of $\theta$ because it is easier for me) To compute the integral in the unit circle, you need to consider not $r=1$ but $r<1$ and $0<t<2\pi$ (they should have taught you this, It would be very difficult for me to explain this here using only text). So our integral is:

$$\int^{2\pi}_0\int^1_0\ln(1+r^2)r\,dr\,dt$$

Substitute $u=r^2$ $\frac{du}{2}=rdr$ and we will get:

$$\int^{2\pi}_0 \frac12 \int^1_0\ln(1+u)\,du\,dt$$

If you integrate by parts, you'll find that the antiderivative of $\ln(u+1)=(u+1) \ln (u+1)-u$, and the whole integral turns out to be:

$$2\pi\ln2-\pi$$

$b)$ I will assume we will find the surface of the Lemniscate. We can divide our function into 4 equal parts, only compute it for the first quadrant, and multiply by 4.

We start by inserting $x=r \cos t \, ,y=r \sin t$ into the equation of the Lemniscate. We will get:

$$r^4=4r^2(\cos ^2t-\sin ^2 t)=4r^2\cos (2t)$$

We want the positive $r$ value so we have $r=2\sqrt{\cos(2t)}$. Hence, $r$ goes from $0$ to $r=2\sqrt{\cos(2t)}$.

We also need the maximum $t$ value. We will need to find the angle of this tangent line:

enter image description here

To find it, we will use implicit derivative of the lemiscate evaluated at $(x,y)=(0,0)$. Which turns out to be $\pm1$. We are concerned with the first quadrant, so we take $+1$. The angle that makes the slope $+1$ is $\arctan(1)=\pi/4$

The surface integral is:

$$4\int^{\pi/4}_0\int^{2\sqrt{\cos(2t)}}_0 r\,dr\,dt$$

I can assume you can take it form here? The answer should come out to be $4$

Thanks for the fast response. I will try this out once I'm home. — Konstantin, May 25 '15 at 09:36
I think you are calculating the area enclosed by the lemniscate, which is presumably what the instructor intended. — user84413, May 25 '15 at 16:00

Double integral (and for the area enclosed by a lemniscate).

1 Answers1