Guys im trying to figure out calculus so as an beginner level question I wanted to calculate the area of any circle with radius r.
I did this in two ways, one way worked, the other one didnt, and I have no idea why not.
Bascially I did the following in the first one: I split the circle into smaller circles, then I calculated all the Circumfrences of all dr's and summed them up. With that method I get the Integral over 2*Pi*r dr which of course solves to Pi*r^2.
This is A in the attached image.
In B I first split the circle into 4 quadrants, then I tried defining the relations of all the little rectangles that are dx * y, I noticed that y is variable, and a function of x, so I drew out what the function must be by using Theta, it turned out to be y(x) = sin(arccos(x/r)
Then I took the Integral of these y(x) * dx from x=0 to x=r, which is one quadrant of the circle. This evaluated to (Pi*r)/4, multiplying by 4 should give me the Area of all 4 Quadrants combined, so Pi*r.
This is as we know not correct, its supposed to be Pi*r^2, im just so confused to where exactly I am missing out on the second r, since the method itself doesnt look that flawed to me.
Many thanks in advance!
Might be valueable:
I evaluated the Integral in Method B from 0 to r, as the quadrant only goes from x=0 to x=r
>>8952028
make wedges idiot
Shouldnt y(x)=r sin (theta)
>>8952039
true! fml
that might be the solution, second
>>8952044
yup, that evaluates to Pi*r^2
Where do I have my head sometimes
>>8952028
>>8952049
What you said about theta = arcos (x/r) is true but when you integrate over x, you're skipping some of the quadrant. As x increases, the distance between x and the point on the circumference decreases.
So what the other guy said works.
Keep what you said in mind. It's handy for fluid mechanics and electromagnetism.
>>8952044
lmao