Force due to fluid pressure

**VitaX** · January 27th 2010, 07:55 PM

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?

**VonNemo19** · January 27th 2010, 08:38 PM

Originally Posted by VitaX

so am I supposed to change the form of the given equation into terms of y instead of x?

Yes

**VitaX** · January 27th 2010, 09:06 PM

Originally Posted by VonNemo19

Yes

How is my work thus far in the problem?

**VonNemo19** · January 27th 2010, 09:13 PM

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 $h=6-y$ ?

**VitaX** · January 27th 2010, 09:25 PM

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.

**VonNemo19** · January 27th 2010, 09:29 PM

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.

**VitaX** · January 27th 2010, 09:30 PM

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?

**VonNemo19** · January 27th 2010, 09:38 PM

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?

**VitaX** · January 27th 2010, 09:50 PM

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?

**VonNemo19** · January 27th 2010, 09:59 PM

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

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

**VitaX** · January 27th 2010, 10:43 PM

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

Is this correct so far?

Crap accidently did a double post when editing an error

**VitaX** · January 27th 2010, 10:55 PM

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

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

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