Area of parametrized curve

Question

A simple closed curve $C$ is given by $\textbf{x} = (f(t), tf(t)), t\in [a,b]$.

Show that the area enclosed by the curve is given by $A = \frac{1}{2}\int_{a}^{b}f(t)^2dt$.

I tried to use Green's theorem here, so choosing $\mathbf{F} = (0, -x)$ implies that the area is given by $$\int_{C}^{} \mathbf{F} \cdot \mathbf{dx} = \int_a^b -f(t)\ (\frac{d}{dt}tf(t))\ dt = \int_a^b -f(t)(f(t) + tf'(t))\ dt = \int_a^b -f(t)^2 - tf(t)f'(t)\ dt$$

Then integration by parts gives me that $$\int_a^b tf(t)f'(t) = \frac{1}{2}(tf(t)^2 - \int_a^b f(t)^2\ dt)$$ and this result into the original yields

$$ \int_C \mathbf{F} \cdot \mathbf{dx} = -\frac{1}{2}\int_a^b f(t)^2\ dt + tf(t)^2$$

I'm not sure where I'm going wrong or if this is the right approach to showing this result. How would I move forward here?

Answer 1

Let $F = (M, N)$, then, by greens theorem, if $\partial \Sigma$ is piecewise smooth : $$ \oint_{\partial \Sigma} \textbf{F} .d\textbf{r} = \iint_{\Sigma}\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dA $$ We are given that $\partial \Sigma$ is described by $\textbf{r} = (f(t), tf(t))$ where $t \in [a, b]$. If we let $M = -y$ and $N = x$, we get: $$ \int_{a}^{b} (-tf(t), f(t)).(f'(t), f(t) + tf'(t))dt = \iint_{\Sigma} 1 - (-1)dA $$ $$ \int_{a}^{b} -tf(t)f'(t)+f^2(t)+tf(t)f'(t)dt = 2\iint_{\Sigma}dA $$ $$ \int_{a}^{b} f^2(t)dt = 2\iint_{\Sigma}dA $$ $$ \frac{1}{2}\int_{a}^{b} f^2(t)dt = \iint_{\Sigma}dA $$ $$ \therefore A = \frac{1}{2}\int_{a}^{b} f^2(t)dt $$

Answer 2

Parametric curves are just screaming out to be solved in the complex plane. Consider that

$$ z=f(t)+itf(t)\\ A=\frac{1}{2}\int\mathfrak{Im}\{z^* \dot z\}\ dt $$ $$ z^*=f-itf\\ \dot z=\dot f+i(t\dot f+f)\\ z^*\dot z=(f-itf)[\dot f+i(t\dot f+f)]=f\dot f+(t\dot f+f)+i[f(t\dot f+f)-tf\dot f]\\ \mathfrak{Im}\{z^* \dot z\}=f^2 $$

Therefore

$$A=\frac{1}{2}\int_a^b f^2(t)\,dt$$