areaintegrationpolar coordinates
can anyone help me calculate this area? I have to use double integrals, and the question sounds like this:
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Calculate the area bounded by the curve $(x^2+y^2)^2=a^2(x^2-y^2)$, where $a$ is a real constant. I have searched online and found that this type of curve is a lemniscate, but I do not know how to convert $x,y$ to polar coordinates.
Could you please point me in the right direction ?
$a)$ 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:
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 $4$
If I understand you correctly, you would like to have a discriminant test for polar coordinates, so that you can easily see whether the critical point of a function $f(r,\theta)$ is an extremum or a saddle point.
The transformation of the discriminant to polar coordinates is a straightforward if somewhat tedious exercise. A direct calculation spawns a great number of terms, most of which conspire to annihilate one another. The final result can then be written as the determinant of a symmetric matrix, $$ \Delta(f) = \det\left[ \begin{array}{cc} \frac{\partial^2f}{\partial r^2} & \frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta} -\frac{1}{r^2}\frac{\partial f}{\partial\theta}\\ \frac{1}{r}\frac{\partial^2f}{\partial \theta\partial r} -\frac{1}{r^2}\frac{\partial f}{\partial\theta} & \frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2} +\frac{1}{r}\frac{\partial f}{\partial r} \end{array} \right] $$
Note that the individual entries in the above matrix are ${\it not}$ equal to the corresponding entries in the cartesian version, but the result for the determinant is the same. In particular, for any twice differentiable function $F(x,y)$, if you feed $$ f(r,\theta) = F(r\cos\theta,r\sin\theta) $$ into the above expression for $\Delta(f)$ and simplify, the result will be $$ \Delta(f) = F_{xx}(r\cos\theta,r\sin\theta)F_{yy}(r\cos\theta,r\sin\theta)-\left[F_{xy}(r\cos\theta,r\sin\theta)\right]^2, $$ just as you would wish.
Best Answer
Since the lemniscate encloses 4 equal subregions, one in each quadrant, you could use
$\displaystyle A=4\int_0^{\frac{\pi}{4}}\int_0^{a\sqrt{\cos2\theta}}r\; dr d\theta$.