# Math Help - Force due to fluid pressure

1. ## Force due to fluid pressure

A window on the side of a 5 foot deep swimming pool is in the shape of an inverted parabola $y=\frac{(9-x^2)}{3}$ located at the bottom of the pool 2 feet below the surface. Find the force F of water on this window.

This formula is used: $F=\int_a^b whA$ $dy$ where w is the density of the fluid, h is the strip depth and A is $L(y)dy$.

$x =\sqrt{9 - 3y}$

$F=\int_0^3 62.4(5-y)(2\sqrt{9-3y})$ $dy$

$F= 124.8\left[\int_0^3 5\sqrt{9-3y} - \int_0^3 y\sqrt{9-3y}\right]$ $dy$

I'm not sure if its supposed to be $2\sqrt{9-3y}$ or just $\sqrt{9-3y}$ if someone can answer this part than I should be able to do the rest. However my friend says the first is u substitution but the 2nd one is integration by parts. What exactly is integration by parts?

2. Originally Posted by VitaX
so am I supposed to change the form of the given equation into terms of y instead of x?
Yes

3. Originally Posted by VonNemo19
Yes
How is my work thus far in the problem?

4. Originally Posted by VitaX
How is my work thus far in the problem?
Yes, it should be $2\sqrt{9-x^2}$, but where are you getting $h=6-y$?

5. Originally Posted by VonNemo19
Yes, it should be $2\sqrt{9-x^2}$, but where are you getting $h=6-y$?
Well in all my notes when the object is submerged in water and total height was taken to be from the height of the surface in this case 6 subtracting the height of an unknown strip length in the object which is y.

6. But, this is what your post says:

Originally Posted by VitaX
A window on the side of a 5 foot deep swimming pool...
I was just clarifying. You've got the idea. If it's 6, then use 6.

7. Originally Posted by VonNemo19
But, this is what your post says:

I was just clarifying. You've got the idea. If it's 6, then use 6.
oh crap you are correct. I dont know where I got 6 from. Now that I've set the integral up correctly. Could you help me integrate this, I'm a little lost on the 2nd part, first part is clearly u sub right?

8. Originally Posted by VitaX
oh crap you are correct. I dont know where I got 6 from. Now that I've set the integral up correctly. Could you help me integrate this, I'm a little lost on the 2nd part, first part is clearly u sub right?
$u=\sqrt{9-3y}$ imples that $y=\frac{9-u^2}{3}$ so that $dy=-\frac2{3}udu$

Can you see what to do?

9. Originally Posted by VonNemo19
$u=\sqrt{9-3y}$ imples that $y=\frac{9-u^2}{3}$ so that $dy=-\frac2{3}udu$

Can you see what to do?
Hmmmm I see that you let $u=\sqrt{9-3y}$ so that then you can say $u=x$ and therefore sub u for x in the original function. But I dont really see how you found what dy is equal to and say I did understand this part, what would the next step be in the integral?

10. Originally Posted by VitaX
Hmmmm I see that you let $u=\sqrt{9-3y}$ so that then you can say $u=x$ and therefore sub u for x in the original function. But I dont really see how you found what dy is equal to and say I did understand this part, what would the next step be in the integral?
Well, one thing at a time. I want you to understand what I've done.

Let $u=\sqrt{9-3y}$

Square both sides

$u^2=9-3y$

Solve for y

$y=\frac{9-u^2}{3}$

differentiate

$dy=-\frac{2}{3}udu$

Now substitute

$\int_0^3y\sqrt{9-3y}dy=\int_{y=0}^{y=3}\overbrace{\frac{9-u^2}{3}}^y\overbrace{(u)}^{\sqrt{9-3y}}\overbrace{\left(\frac{-2}{3}udu\right)}^{dy}$

11. $F = 124.8\left[\int_3^0 5u\left(\frac{-2}{3}u\right)\right - \int_3^0 \frac{9-u^2}{3}u\left(\frac{-2}{3}u\right)\right]$ $du$

Is this correct so far?

Crap accidently did a double post when editing an error

12. $F = 124.8\left[\int_3^0 5u - \int_3^0 \frac{9-u^2}{3}(u)\left(\frac{-2}{3}u\right)\right]$ $du$

Is this correct so far? Or is the $\frac{-2}{3}u$ supposed to be in the first integral as well