Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart

Question

Hi all I am trying to figure out how to sketch the heart. Here is what I have tried so far: $$r = 1 - \cos(\theta) \\ r(r = 1 - \cos(\theta)) \\ r^2 = r - r\cos(\theta) \\ $$ Use the fact that $$r =\sqrt{x^2 + y^2} \text{ and } x=r\cos(\theta) \text{ to rewrite the equation as , } \\ x^2 + y^2 = \sqrt{x^2 + y^2} + x \\ x^2 - 2x + y^2 - y = 0 \\ \text{Completing the square gives } \\ (x-1)^2 + (y-\frac{1}{2})^2 = -\frac{5}{4}\\ \sqrt{(x-1)^2 + (y-\frac{1}{2})^2 } = \sqrt{\frac{5}{4}} \\ (x-1) + (y-\frac{1}{2}) = \sqrt{\frac{5}{4}} \\ (y-\frac{1}{2}) = -(x-1) + \sqrt{\frac{5}{4}} \\ y = -(x-1) + \sqrt{\frac{5}{4}} + \frac{1}{2} \\ y = -x + \frac{\sqrt{5} + 3}{2} $$From here we can get a table of values and sketch the function. For the area the formula is $$\frac{1}{2} \int f(\theta)^2 \,d\theta$$ So should it be $$\frac{1}{2} \int y = -x + \frac{\sqrt{5} + 3}{2}$$

Answer 1

You can easily draw the shape by substituting convenient values for $\theta$. To find the area, use the formula $\displaystyle A=\frac{1}{2}\int_a^{b}r(\theta)^2d\theta$. The integration proceeds as follows:

$$2\cdot \frac{1}{2} \int_0^{\pi}(1-\cos \theta)^2d\theta$$

$$ \int_0^{\pi}(1-2\cos \theta+\cos^2\theta )d\theta$$

$$ \int_0^{\pi}1-2\cos \theta+\frac{1}{2}+\frac{1}{2}\cos 2 x )d\theta$$

$$ \int_0^{\pi}\frac{3}{2}-2\cos \theta+\frac{1}{2}\cos 2 x )d\theta$$

$$ (\frac{3}{2}\theta-2\sin \theta+\frac{1}{4}\sin 2 x) \Bigg|_0^{\pi}$$

$$\frac{3}{2}\pi-0+0-(0-0+0)=\frac{3}{2}\pi$$

Answer 2

Drawing the figure is the hard part by far. For example, Matlab doesn't have a method for directly adding arrows to a curve like this. What I tried was to draw a little right-pointing arrowhead with its tip at $(0,0)$ and some reasonable size. Then at a selection of points along the path, I rotated that arrow counterclockwise by $\text{atan2}\left(\frac{dy}{d\theta},\frac{dx}{d\theta}\right)$ and then added $(x(\theta),y(\theta))$ to its coordinates. Here is my Matlab code:

% cardioid.m

% Plot main curve
theta = linspace(0,2*pi,500);
r = 1-cos(theta);
x = r.*cos(theta);
y = r.*sin(theta);
plot(x,y,'b-');
hold on;
% Plot arrows
asize = 0.07;
axy = [-1 0 -1;
    -1 0 1]*asize/sqrt(2);
% We want fewer points
theta = linspace(0,2*pi,10);
% Avoid the cusp
theta = theta(2:end-1);
r = 1-cos(theta);
x = r.*cos(theta);
y = r.*sin(theta);
% Orientations needed
drdtheta = sin(theta);
phi = atan2(sin(theta).*drdtheta+r.*cos(theta), ...
    cos(theta).*drdtheta-r.*sin(theta));
for k = 1:length(theta),
    R = [cos(phi(k)) -sin(phi(k));
        sin(phi(k)) cos(phi(k))];
    xy = R*axy;
    plot(xy(1,:)+x(k), xy(2,:)+y(k),'b-');
end
% Titles and labels
axis equal;
xlabel('x');
ylabel('y');
title('Cardioid with Direction');
hold off;

And here is the figure it drew:

Having drawn the cardioid, finding its area is anticlimactic: $$A=\int_0^{2\pi}\int_0^{1-\cos\theta}r\,dr\,d\theta=\int_0^{2\pi}\frac12\left(1-2\cos\theta+\cos^2\theta\right)d\theta=\frac12(2\pi)\left(1-0+\frac12\right)=\frac{3\pi}2$$ Where we have used the average values for $1$ of $1$, for $\cos\theta$ of $0$ and for $\cos^2\theta$ of $\frac12$ over the interval of length $2\pi$ to zip through the azimuthal integral.