Find the area between two curves:
- $x \ge 0$
- $(x^2+y^2)^2=x^2-y^2$
using double integrals.
Finding area between two curves using double integral
calculus
Related Solutions
You'll want to find the point of intersection, and use the $t=x$-value of this point as a bound for integrating (see note below).
Set the two equations equal to one another, and solve for the values $t$ (the x-coordinate) that solves the resulting equation: there will be one such value on your domain.
$$9\sin(t) = 10\cos(t)\implies \frac{\sin t}{\cos t} = \frac {10}{9}\implies \tan t = \frac{10}{9} \implies t = \tan^{-1} \frac{10}{9}$$
Then calculate the sum of the integrals $$\int_0^t (10\cos x - 9\sin x)\,dx + \int_t^\pi (9\sin x- 10\cos x)\,dx.$$
Note: you need to know the point of intersection to divide the integral into two parts, because from $0\leq x \lt t$, where t is the value of x at the point of intersection, $10\cos x > 9\sin x$. At $x = t$, $10\cos x = 9 \sin x$. And from $t \lt x \leq \pi$, $9\sin x > 10 \cos x$. So to measure total area, we need to ensure we choose the correct function as the "upper bound" of each region when we integrate to obtain the total area of the regions bound by the functions.
There isn't much difference between doing area integration in polar coordinates as a double integral and in the way you may have encountered it earlier in single-variable calculus. It is still important to have an idea of what the regions look like (here, you have a limacon and a "peanut").
Your radial portion is correct, but you really need to look at the angles where the curves intersect: you'll want to solve $ \ 2 + \sin \theta \ = \ 2 + \cos (2 \theta)$ to get the range of angle integration. There are two zones to cover, but you can make use of symmetry here and just integrate over one of them.
The red curve is the limacon $2 + \sin\theta$ , the blue curve, $2 + \cos(2 \theta)$ .
Incidentally, in the integral you wrote, it's not that there is no integrand, but rather that the integrand is "1". You are doing a surface integral where all infinitesimal patches are equally "weighted" at 1 , so the result is simply the area of the region. Surface integrals with some "weighting function" $f(r,\theta)$ or $g(x,y)$ turn up in various applications.
Best Answer
Convert to polar coordinates
$$ r= \sqrt{\cos 2 \theta}$$
Area in first quadrant
$$=\int_0 ^{\pi/4} r^2/2\; d\theta = \int \frac12 {\cos 2 \theta} \;d\theta =\dfrac{\sin 2 \theta}{4}|^{\pi/4}_0 = \frac 14 $$
For $x\ge0$ area is to be doubled.