I've been wondering if there was another way of calculating the area of a circle by integration, other than using trig substitution. So I decided to try dividing the circle into many rectangles, and essentially adding them all up, kind of like here, but doubled . I changed the circle equation to solve for y: $$ x^2+y^2=1\\y=\sqrt{1-x^2}\\2y=2\sqrt{1-x^2} $$ And used that to find my rectangle height at any point on the x-axis (The 2y is for the full rectangle, not just the half above the x-axis.) The base of the rectangle is $dx$, and so the area should be $$ dA=2y\ dx $$ Adding them all up should result in $$ A=\int_{-1}^{1}2y\ dx\\A=2\int_{-1}^1\sqrt{1-x^2}\ dx $$ Now I think this is correct, and I tried integrating and didn't give me any good answer. I know the answer is $\pi$, because we all now that $A=\pi r^2$ and $r=1$. Even then, I know something is wrong with what I am trying to do, because I can't find something that will lead to a $\pi$ as an answer. Is it just impossible to find the area of circle with "rectangulation" (lol), or is it just me that is not doing something proprely? I hope it's me that's wrong! Thanks
P.S: It is very late and I have been working on that for quite a long time, so it's is totally possible that I'm just not really seeing things right.
What makes you think it is incorrect? Do not forget that:
$$2\cdot\int\sqrt{1-x^2}=x\sqrt{1-x^2}+\sin^{-1}x$$
Have a look here if you're forgotten the derivation.