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I know how to do this problem, but I'm stuck at the arc length differential.

Set up an integral for the arc length of the curve. (Do not evaluate the integral)

x=y^2ln(y), 1<y<2
dx/dy = 2yln(y) + y

ds= sqrt (1 + (2yln(y)+y)^2
so far I have sqrt (1 + 4y^2ln(y) + 2[2y^2ln(y) + y^2) dy
I'm stuck at this part

1 answer

ds=sqrt(dx^2+dy^2)

dx=(2ylny +y^2/y )dy

S=INT ds=INT ((2ylny+y)^2 + 1)dy from y=1 to 2

check that
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