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Calculate second moment of area?
Calculate the second moment of area with respect to the y-axis for a circle of radius R = 1, centered at (x,y) = (1,2).
been stuck on it for 2 hours 40 mins running. don't laugh :/
sorry i'm confused with some of the description in your working. let me picture the integral correctly
the inner integral: [(x^2) / {(x-1)^2 + (y-2)^2}] with lower limit 2-sqrt[1-(x-1)^2] and upper limit 2+sqrt[1-(1-x)^2] with respect to y
the outer integral has lower limit 0 and upper limit 2, respect to x
is that right? how did you arrive at that in the first place? there's nothing like this in my text book or notes. the only example shown is
x^2 + y^2 < (or equal to) R^2
then second moment of area with respect to x is
I = integral y^2 dx dy = (R^4)pi/4
1 Answer
- 1 decade agoFavorite Answer
I didn't find any simple way to do this. It is the double integral of x^2 over the circle. This is the iterated integral with outer limits in x of 0 to 2 and inner limits in y of 2-sqrt[1-(x-1)^2] to 2+sqrt[1-(1-x)^2]. The inner integral is trivial, so it becomes the single integral of x between 0 to 2 of 2 x^2 sqrt[1-(x-1)^2].
Let u = x-1 to get the integral of u between -1 and 1 of 2(u+1)^2 sqrt(1-u^2). Multiply out the (u+1)^2 to get three integrals, the middle of which is 0 since it is an odd function. The other two are even, so the integral is 2 times the value between 0 and 1. Use standard half angle theorems to finish these two integrals using a trig substitution, u = sin theta.