- #1

DunWorry

- 40

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## Homework Statement

Find the area in the polar curve r = sin2θ between 0 and [itex]\frac{\pi}{2}[/itex].

The way to do this is to say the area of a tiny bit of this polar curve, dA = [itex]\frac{1}{2}[/itex]r[itex]^{2}[/itex]dθ

so the integral is just [itex]\frac{1}{2}[/itex][itex]\int[/itex][itex]^{\frac{\pi}{2}}_{0}[/itex](sin2θ)[itex]^{2}[/itex]dθ

if we did say a function in cartesian coordinates, eg y=x we just do [itex]\int[/itex]x dx. I am confused as to why in polar coordinates we cannot just do the same and say do [itex]\int[/itex]sin2θ dθ, I understand how to get the correct integral, but what I am asking is what is wrong with just integrating the polar function like we do in cartesian coordinates? why do we have to say we will take a bit of this sector and integrate it over these limits, for polar function, but for a cartesian one we just plug the function into the integral?

Thanks