Area of Cardioid $r=1+\sin(\theta)$ Using Green's Theorem

Question

Find the area enclosed by $r=1+\sin \theta$ using Green's theorem.

What I have is

$\gamma(t)=(t,1+\sin t)$

$\gamma'(t)=(1,\cos t)$

Then $\frac{1}{2}\int_{0}^{2\pi}(-1-\sin t+t\cos t)dt=-\pi$

Is it correct?

Answer 1

The form of Green's theorem that you used: $$ A(D) = \iint_D dx\,dy = \frac{1}{2}\oint_{\partial D} (x\,dy - y\,dx) $$ applies when $x$ and $y$ are cartesian coordinates. Your expression $\gamma(t) = (t,1+\sin t)$ expresses the curve in polar coordinates. So you need to either state Green's theorem in polar coordinates, or express $\gamma$ in cartesian coordinates.

Let's do the latter. The change of variables is $x = r\cos\theta$, $y = r\sin\theta$. So use $$ \gamma(t) = ((1+\sin t)\cos t,(1+\sin t)\sin t) $$

As for the former, you can compute \begin{align*} dx &= \cos\theta \,dr - r \sin\theta \,d\theta \\ dy &= \sin\theta \,dr + r \cos\theta \,d\theta \end{align*} Then through a little exterior algebra, we get $$ dx \,dy = r\,dr\,d\theta $$ and $$ x\,dy - y\,dx = r^2\,d\theta $$ So Green's theorem tells us $$ A(D) = \int_D r\,dr\,d\theta = \frac{1}{2}\oint_{\partial D} r^2\,d\theta $$ I think in many undergraduate multivariable calculus courses this identity isn't derived from Green's theorem, but it can be. In any case, if $D$ is the region enclosed by the cardioid $r = 1 + \sin\theta$, then $$ A(D) = \frac{1}{2}\int_{0}^{2\pi}(1+\sin\theta)^2\,d\theta $$

Answer 2

While I think Matthew Leingang work is more elegant and more general. Here is a slightly different approach.

Green's Theorem: $\iint (\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y})\ dx\ dy = \oint P dx + Q dy$

Choose $P, Q$ such that $(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}) = 1.$ We have a wide range of options here. The point is to choose the easiest to work with. e.g. Let $P,Q = 0,x$

$r = 1 + \sin \theta\\ x = r\cos \theta = (1+\sin\theta)\cos\theta\\ dx = -\sin\theta + \cos^2\theta - \sin^2\theta\ d\theta\\ y = r\sin \theta = (1+\sin\theta)\sin\theta\\ dy = \cos\theta + 2\sin\theta\cos\theta\ d\theta$

$\oint x\ dy = \int_0^{2\pi} (\cos\theta + \frac 12 \sin 2\theta)(\cos\theta + \sin 2\theta)\ d\theta = \frac {3\pi}{2}$