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Find the work done in pumping the water out of the top of a cylindrical tank 3.00 ft in radius and 10.0 ft high, given that the tank is initially full and water weighs $62.4 \mathrm{lb} / \mathrm{ft}^{3}$. [Hint: If horizontal slices $d x$ ft thick are used, each element weighs $62.4(\pi)\left(3.00^{2}\right) d x$ lb, and each element must be raised $10-x \mathrm{ft}$, if $x$ is the distance from the base to the element (see Fig. 26.66 ). In this way, the force, which is the weight of the slice, and the distance through which the force acts are determined. Thus, the products of force and distance are summed by integration.]

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 6

Other Applications

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so the card a initially as floats the cylinders and static equilibrium. So we know that the buoyancy force would be equaling the gravitational force. And this would equaling the density of the liquid multiplied by the cross sectional area multiplied by the height multiplied by the acceleration due to gravity. A an external force f pushes, then down on the cylinder, a distance of axe. And so now we have a submerged the length and so the buoyancy force would be equaling the gravitational force, plus this external force that it's acting on the top of the cylinder Now. This would be equaling that the density of the liquid multiplied by the cross sectional area of the cylinder, multiplied by H plus ax multiplied by G. And so we know that the density of liquid multiplied by a H G is equal in the gravitational force. So now we're left with the external force Equalling the density of the liquid multiplied by the cross sectional area multiplied by G. The acceleration due to gravity times X, the additional distance with additional depth that the cylinder is pushed down into a liquid for part B. Since the amount of work done Ah Would be done by the external force to push the cylinder from X two X plus a, uh, infinite testimony. Small additional distance. Essentially the differential work, the infinite testimony. Small work would then be equaling the force multiplied by the infinite testimony. Small distance, uh, over which the force is applied. This would be equaling the density of the liquid multiplied by a G axe again, DX and so So push the cylinder from zero meters to 10 centimeters. It requires work integrating from the initial X to the final x of the force DX. This is then Equalling to the density of the liquid multiplied by okay, multiplied by G. These were all these air, all constants integrating from zero 2.10 meters x dx. And so we have then the work equaling one house times the density of the liquid multiplied by a G on X initial squared minus X final squared and this is evaluated at zero and at 00.10 meters. And so we have. Then the work Equalling 1/2 multiplied by 10 to the third kilograms per cubic meter multiplied by point 0 to 0 meters quantity squared multiplied by pi multiplied by 9.8 meters per second, squared multiplied by 0.10 meters, quantity squared and the total work done pushing the cylinder would be 0.62 jewels. That is the end of the solution. Thank you for watching.

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