# Thread: Bisection method problem #2(check answer)

1. ## Bisection method problem #2(check answer)

Hello everybody. I'm getting headache of not getting the right result.

A trough of length $L$ has a cross section in the shape of a semicircle with radius $r$. When filled with water to within a distance $h$ of the top, the volume $V$ of water is $V = L[0.5\pi r^2-r^2\arcsin{(h/r)}-h(r^2-h^2)^{1/2}]$

Suppose $L=10$ ft, $r=1$ ft, and $V=12.4$ $ft^3$. Find the depth of water in the trough to within 0.01 ft.
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Now I'm supposed to find $h$ from this equation

$10[0.5\pi-\arcsin h - h(1-h^2)^{1/2}]=12.4$

and my solution is approx. 0.16(and I know it's right for the above equation) while the answer in book is 'the depth of water is 0.838 ft.'

2. Your answer is correct and easily verifiable.

3. hihihi...long time since I posted this problem. I solved it after I reviewed the figure again and I got 10 out of 10 in num analysis . Good to know there are people like you who answer these unanswered threads!