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semicircle problem

 imagine the top half of a circle. the origin lies along the bottom of the semicircle, and in the middle. y axis up, and x axis to the right and left. i think theta can only go from 0 to 180 degrees since it is a semi circle. Y = d(theta) R squared R = radius, integrate from 0 to R
You haven't stated a problem here! What do you want to integrate from 0 to R? What function are you integrating?

If you just want the area, look at the problem you gave:
 in a rectangles case, dA = dX dY if i have the rectangles width and it is 4 m, then dA = 4dY
(And arildno did not say he had never seen that before, he was trying to get you to think about why you think that was true.)

Suppose your semi-circles radius is R. Then dA= r dr dθ. In a sense, dr here, as well as r, is r: along each radius we go from r= 0 to R so the "change in r", dr, is R.

Okay, then r dr dθ becomes dA= R2 dθ, just like your dX dY became 4dy, the area of the semi-circle is
$$\int_{\theta=0}^{\pi} R^2 d\theta= \pi R^2$$, exactly the right answer.

(θ does not go from 0 to 180! As I am sure you learned in when you were learning the derivatives of sine and cosine, in calculus, all angles are in radians.)