
Hydrostatic Force
"A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9ft. If the pool is full of water, find the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool."
That is the question and I need help with part C and D. the formula for hydrostatic force is integral from a to b of the weight density (62.5 for water for this problem)*depth*length of the cross section
for part C i split the integral into two pieces since their is rectangle and a triangle in the picture.
i came up with the integral from 0 to 3 (62.5)(40)xdx+ integral from 3 to 9 (62.5)(40/6x)(x)dx
but the answer is suppose to be 4.88x10^4 but i am not close.
can someone help!?!? and im lost on part d

Re: Hydrostatic Force
can someone just help me with part d! i have the rest figured out!!! pleaseeee

Re: Hydrostatic Force
For part c) I would use:
$\displaystyle F=62.5\left(\int_0^3 x(400)\,dx+\int_3^9 x\left(\frac{20}{3}(x3)+40 \right)\,dx \right)$
$\displaystyle F=62.5\left(40\int_0^3 x\,dx+\frac{20}{3}\int_3^9 9xx^2\,dx \right)$
$\displaystyle F=1250\left(\left[x^2 \right]_0^3+\frac{1}{18}\left[27x^22x^3 \right]_3^9 \right)$
$\displaystyle F=1250(9+30)=48750$
For part d) I will try to get you started...
Take the $\displaystyle x$axis coinciding with the slanting bottom as seen from the trapezoidal sides, and express the depth in terms of $\displaystyle x$.