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i have been told that dA=R dR d(theta), what does mean exactly in terms of pi. the shape is a semicircle. Does this mean that d(theta) is 180, or pi. Please give me an example.
You do know how to integrate, don't you?n05tr4d4177u5 said:here is an example of mine that i have
suppose i hae a rectangle,
in a rectangles case, dA = dX dY
if i have the rectangles width and it is 4 m, then dA = 4dY,
which lets me integrate along y axis,
now for a semicircle, how would i do that if dA = r dr d(theta)
Well, why do you believe this to be true?n05tr4d4177u5 said:in a rectangles case, dA = dX dY
if i have the rectangles width and it is 4 m, then dA = 4dY,
You are good up to here.n05tr4d4177u5 said:here is an example of mine that i have
suppose i hae a rectangle,
in a rectangles case, dA = dX dY
Why can you write this? Where did it come from? You cannot just drop the dx and it is not automaticlly set to 4 or any other value.if i have the rectangles width and it is 4 m, then dA = 4dY,
Can you complete the above integral?which lets me integrate along y axis,
Ok, now lets set up the integral for the area of a semi circle.now for a semicircle, how would i do that if dA = r dr d(theta)
You haven't stated a problem here! What do you want to integrate from 0 to R? What function are you integrating?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
(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.)in a rectangles case, dA = dX dY
if i have the rectangles width and it is 4 m, then dA = 4dY