Hydrostatic force on water tank problem

[Gauss177]

Homework Statement

The end of a tank containing water is vertical and has the indicated shape (in attached picture). Explain how to approximate the hydrostatic force against the end of the tank by a Riemann sum. Then express the force as an integral and evaluate it.

Homework Equations

P=1000gx
F=P*A (pressure * area)

The Attempt at a Solution

I think my main problem is finding the area of the ith strip. The pressure is relatively easy to calculate, and if I know the pressure and area I can integrate to find the hydrostatic force.

 

[AlephZero]

Does the fact that the picture says "10 m (radius)" mean the end of the tank is a semicircle?

If so, you can use the equation of the circle, x^2 + y^2 = r^2, to find the area of each strip.

If it is some other shape, the question should tell you what the shape is.

 

[Gauss177]

Yes, the end of the tank is a semicircle. Looking at a similar example in the book, is the length of each strip 2 \sqrt{100-y_i^2}? Because in the book's example, the end of the tank is a full circle with radius 3, and they got the length of each strip to be 2 \sqrt{9-y_i^2}. I'm not sure how to get this though from the equation of the circle.

Thanks

 

[AlephZero]

The equation of the circle is x^2 + y^2 = 10^2

So x = \pm\sqrt{100-y^2}

The strip at height y_i goes from x = -\sqrt{100-y_i^2} to x = +\sqrt{100-y_i^2} so its length is 2\sqrt{100-y_i^2}

 

[Gauss177]

Thanks. Can you quickly check if I did this right:

\displaystyle{\int_{0}^{5} 62.5y(2 \sqrt{100-y}) \;dy}
\displaystyle{125 \int_{0}^{5} y \sqrt{100-y} \;dy}

Substituting u=100-y, du=-dy:

\displaystyle{-125 \int_{100}^{95} (u+100) \sqrt{u} \;du}

I get the answer to be 1218880 lb for hydrostatic force. This seems really big, so did I do something wrong above?