I was wondering what i am doing wrong in my proof of the formula $$A=\int r^2/2$$ First of all we know, given parametric equations $x=f(t)$ and $y=h(t)$ the area represented by that parametric curve is $$\int h(t)*f'(t)$$ (A side not question about that, will this "double" count the area if the parametric curve makes loops in the same place? Also is this actual area, or area under the graph?). Now assuming validity of this foprmula we know that apolar function can be seen as a parametric equation $x=r\cos(\theta)$ $y=r\sin(\theta)$. Hence using the formula we obtain $$\int -r^2\sin^2(\theta)+rr'sin(\theta)\cos(\theta)$$ Now unless i am missing something very obious, $r^2/2$ is not equal the expression above. What is going on here?
Answer
The quoted formula, $$\int h(t)*f'(t) dt$$ refers to the area under the graph, down to the $x$-axis. This is because $$\int h(t)f'(t) dt=\int y\frac{dx}{dt}dt=\int y dx.$$ In fact your deduction is correct and your second formula gives the area under a polar curve, down to the $x$-axis.
However, the formula is sensitive to the direction that the curve is following. So if the curve moves towards the negative $x$-direction, then the resulting area is negative (because $f'(t)<0$). As a test, try $r=1$ (circle) from $\theta=0$ to $\theta=\pi/2$; your formula gives the correct $\pi/4$ (but negative).
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