Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart 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
$$\frac{1}{2} \int y = -x + \frac{\sqrt{5} + 3}{2}$$

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:
$$2\cdot \frac{1}{2} \int_0^{\pi}(1-\cos \theta)^2d\theta$$
$$ \int_0^{\pi}(1-2\cos \theta+\cos^2\theta )d\theta$$
$$ \int_0^{\pi}1-2\cos \theta+\frac{1}{2}+\frac{1}{2}\cos 2 x )d\theta$$
$$ \int_0^{\pi}\frac{3}{2}-2\cos \theta+\frac{1}{2}\cos 2 x )d\theta$$
$$ (\frac{3}{2}\theta-2\sin \theta+\frac{1}{4}\sin 2 x) \Bigg|_0^{\pi}$$
$$\frac{3}{2}\pi-0+0-(0-0+0)=\frac{3}{2}\pi$$
