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# A vertical plate is submerged (or partially submerged) in water and has the indicated shape. Explain how to approximate the hydrostatic force against one side of the plate by a Riemann sum.Then express the force as an integral and evaluate it.

## $2.36 \times 10^{7} \mathrm{N}$

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Applications of Integration

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this problem. We're given that there is an object with a circular cross section that is submerged in water. And we know that radius off this circle as eight meters. Let's assume that the center of the circle is our origin sort of 00 And we have then the eggs and the alliances. It's such Now let's assume not, um let's actually write what we know. We know that force is equal to pressure times area and pressures equal to row times g times, step times area. No, uh, we know role. And that is 1000. We nog. That is 9.8 meters per second squared. Now we need to determine detect and the area. Now, as you can see, we have a no Jupiter circular cross section and we know that radius lay. Sorry question of that circle would be excreted. Plus y skirt is equal to our scars. Sort of a squared from this weekend, right, x in terms of why, as X is equal to 64 minus Weisberg square root. Now, if that is ex now assume it. We have a circle. This is the center. Uh So the area would be since you need the area in order to calculate the force area would be the area of distant strip summed up over all this or, um along the diameter off this circle. And this would be, since we just found what excess in terms of why so one side people have skirted off 64 but it's widespread. And since circle is symmetric, we will have the same bling on the other side of 64 minus squared. Unless it's in red, the thickness as D y. Since we assumed why access to be the word to go access so area off the 10 strip would be less assume. Not that that is D A. Meaning that it shows us that does a differential area. It would be a total languages, too. Times Square of 64 1 It's widespread both. Bye bye, d y now let's senate. We are measuring. Why, since this is our origin, this is positive. Why? But don't forget. Pressure depends on the depth and did that. We actually are measuring the duct of from the surface off the water. So if this distance is for and if the radius is eight, which means that total distance from the, um, origin surfaces 12 meters and the death would then be 12th mind swine. So this is D so we know what he is. We know what a is less calculate than the force acting on this very tin plate. That would be DF so DF would be row times g times 12 minus y. This is depth multiplied by inference. Some old this differential area that is to turn screwed off. 641 It's one straight times d y. Now we would find the total force acting on this subject if you work some of those 10 areas up so we could use room on some. So total left would then be limit as then a purchase to Infinity Summation over I from one end. We have wrote Time Z 12 minus y I times to 64. Two times Skirted off Season four minus y I squared times Delta. Why we can write this one as an integral um Now, since the center again is the origin as sensory aliases ate, this would be done. Why is it going to negative eight level and the upper bound will be wise ical to positive eight double, so the integral limits often trigger, will be negative. 88 He had ro times You are Rose. 1000 g is 9.8. He had to. This is a constant again we know had 12 minus y times skirted off 64 minus y squared Do you want? Okay, we can write this one as 19,600 in Trickle Native 8 to 8 12 minus y times Skirted of 64 months. Why spray Do I know we can separate this internal since we had 12 minus y so we can move by 12. Put this and NATO Why with that And write this internalized 19,600. It's just constant. We have internal for many to 8 to 8 this statement. Infinity. Um, we have 12 times skirted of 64 but it's Weiss cried d y minus in triangle from negative 8 to 8. Wine time screwed off 64 minus y squared, Do you I And if you virtually this integral, we would find the total force acting on the side off this object with circular cross section s 2.36 10 37 New chips

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Applications of Integration

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