Solutions - 0100-05

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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Option 1:

(2) Using $y=0$ at water line, increasing downwards: (a) Left:

$$F={\int }_1^5\rho gy\left(2\sqrt{2^2-{(y-3)}^2}\right)dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_0^2\rho gy\left(2\sqrt{4^2-{(2-y)}^2}\right)dy$$

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Option 2:

(2) Using $y=0$ at center of shape, increasing downwards: (a) Left:

$$F={\int }_{-2}^2\rho g(y+3)2\sqrt{2^2-y^2}dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_{-2}^0\rho g(y+2)2\sqrt{4^2-y^2}dy$$