The area Ai will be the product of the width and the thickness of the strip, as indicated below.

• The pressure Pi will be simply `rho g yi, since yi as measured from the water level is the depth.
• The force Fi will be the indicated product of area and pressure.
• The Riemann sum of the force contributions Fi is indicated in the next-to-last line below, where yi takes values between 0 and 20, and the resulting integral in the last line.
• The integral is easily evaluated using the substitution u = 400 - y^2; the result is the total force on the dam.

Video Clip #02

We next determine the velocity required for a small satellite near its surface to completely escape the gravitational attraction of a 1.2 kg pendulum bob of radius 5 cm.

• It is easiest to picture this problem with the spherical bob in empty space, far away from the gravitational influence of the rest of the universe.
• The gravitational attraction of the sphere is quite small, so the velocity required probably will be small.
• To escape the gravitational field of the sphere, the satellite must have a kinetic energy 1/2 m v^2 equal to the work done by the satellite against the gravitational pull of the sphere as it travels from the surface of the sphere to an infinite distance from the sphere.

We begin by calculating the work done against gravity as the satellite moves through distance `drI directly away from the sphere, while at approximate distance ri.

• The work is the product of the force and the distance; the force is determined by the gravitational attraction F = G M m / r^2, where M is the mass of the sphere and m the mass of the satellite.

The work contribution over the interval will be as indicated below.

• The total work will be approximated by a Riemann sum of the Wi contributions, which in the limit gives us the indicated integral for the work Wesc done against gravity by the escaping satellite.
• The integral is easily evaluated, with the indicated result.

Substituting the values of G and the 1.2 kg mass M of the sphere, we obtain Wesc = 1.4 * 10^-9 J / kg * (m), where (m) is the mass of the satellite.

• We set this equal to the kinetic energy KE0 at the surface.
• Using 1/2 (m) v^2 for the kinetic energy, we easily solve for v.
• We might have symbolically solved the equation 1/2 (m) v^2 = G M m / R for v, then substituted G, M and R = 5 cm - .05 m for the resulting expression v = `sqrt( 2 G M / R ).

Video Clip #03

We calculate the kinetic energy of a uniform 50 cm rod rotating with an angular velocity of .5 rad/s about its center.

• We begin by calculating the kinetic energy of the segment of length `dx at position xi as measured from the center of rotation.
• The mass per unit of length will be 40 g / 50 cm, so a segment of length `dx will have mass mi = 40/50 (g / cm) `dxi.
• The velocity of a segment at distance xi from the center of rotation will be the product vi = .5 rad/s * xi of its angular velocity and distance from the center.
• The kinetic energy of the segment is therefore 1/2 mi vi^2, which when we substitute mi and vi simplifies to .1 xi `dxi.
• The total kinetic energy is therefore the Riemann sum of the KEi contributions, with xi ranging from -25 cm to 25 cm (positions on the rod as measured with respect to its center).
• The Riemann sum approaches the indicated integral, which is easily evaluated.

Video Clip #04

We finally calculate the work required to lift the fluid from a uniform cylinder, initially filled to depth 20 cm, to a reservoir 15 cm above the initial level of the fluid.

• We begin by calculating the work required to lift the fluid in a horizontal strip of width `dyi at altitude yi above the bottom of the container.
• The work will be the product of force and distance.
• The force in this case is the weight `rho g Vi of the fluid in the strip, where Vi is the volume of the strip.
• The distance hi is the vertical distance from the strip to the reservoir 15 cm above the initial fluid level.