Mostly, it is stated that area below x axis is negative. For an example, for y=cosx, upon integration(say from -pi to pi), we get area positive above X axis and below X axis, we get negative area, and total area is 0.
But for a circle(having portions below and above x axis), if we move from left side to right side of it (i.e. x=a to x=b),
upon integration it gives us area of upper half only.(e.g. y=sqrt(1-x^2) from x=-1 to 1 gives 1.57 , which is half of the total area.
Is not the second case supposed to give same result like that of first one?
What am I missing?
Update: Similar to cartesian graph, can there be a negative area in polar graph? I have an intuition that going counter clockwise gives positive area while going clockwise gives negative area, is this correct?
Best Answer
In general $$ \int_{a}^bf(x)\, dx $$ gives us the signed area of the region in the plane bounded by the graph of $f$, the $x$-axis and the lines $x=a$, $x=b$. See the picture at the bottom for an illustration.
In particular $$ \int_{-1}^{1}\sqrt{1-x^2}\, dx\tag{0} $$ gives us the signed area bounded by the upper semicircle, the $x$-axis, and the lines $x=1$ and $x=-1$ and hence gives us the area of the upper half disk.
In a similar way, $$ \int_{-1}^{1}-\sqrt{1-x^2}\, dx\tag{1} $$ gives us the signed area bounded by the lower semicircle, the $x$-axis, and the lines $x=1$ and $x=-1$ and hence gives us the negative of the area of the lower half disk.
If we add the two previous integrals together, we get zero.