r = theta ^2
The formula is integral from 0 to pi/4 dt of sqrt(r^2(t) + (r')^2(t)).
Here r(t) = t^2, r'(t) = 2t, the integral isint[sqrt(t^4 + 4t^2)]dt = int[t*sqrt(t^2 + 4)]dt = (1/2)int[sqrt(y + 4)]dy where y = t^2.
Further = (1/2)(2/3)[(y+4)^(3/2)] = (1/3)[(t^2 + 4)^(3/2)] with the substitution t from 0 to pi/4:
(1/3)[ ((pi/4)^2 + 4)^(3/2) - 4^(3/2) ].
The only simplification here is 4^(3/2) = 8.
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