Question

How do I set this up in terms of x? Physics integral.

Answer 1

The pressure pushes normal to the surface and varies linearly with the depth....

Answer 2

Setup:
Pressure = \(\rho g (x-3)\) since first 3 m is not immersed.
Area of strip of height dx and width \(2\sqrt{r^2-x^2}\) = \(2\sqrt{r^2-x^2}dx\)
Force = Pressure \(\times\) area
Total force = \(\int_3^9pressure\times area\)

Answer 3

@mathmate Thanks for the picture, so would the integral look like \[\int\limits_{3}^{9}[pg(x-3)*2\sqrt{r^2-x^2}]dx\]?

Answer 4

Yes, that's what I have too.
But watch out for the answer. It has already factored out \(2\rho g\) outside of the integral sign.